#$&* course Phy201+L 010. Note that there are 10 questions in this set.
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Given Solution: The net force on the block is the product F = m * a of its 10 kg mass and its 2 m/s^2 acceleration. The net force is therefore F = 10 kg * 2 m/s^2 = 20 kg * m / s^2. The unit of force, which is the product of a quantity in kg and another quantity in m/s^2, is just the algebraic product kg * m/s^2 of these two units. This unit, the kg * m / s^2, is called a Newton. So the net force is 20 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. How much force must be exerted by someone pulling on it to accelerate a 10 kg object at 2 m/s^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Assuming that the object is being pulled on a flat surface, then we can use the same equation as before to determine the force (Force = Mass x Acceration). F = 10 kg x 2 m/s/s F = 20 kgm/s/s F = 20 Newtons So to move the object the force will need to exceed a force of 20 Newtons to effectively move the object. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This depends on what forces might be resisting the acceleration of the object. If the object is accelerating on a surface of some type, then there is a good chance that a frictional force is opposing the motion. In this case the person would have to exert more force than if friction was not present. If the object is being pulled upward against the force of gravity, then force must be sufficient to counteract the gravitational force, and in addition to accelerate the object in the upward direction. If the object is being pulled downhill, the force exerted by gravity has a component in the direction of motion. The component of the gravitational force in the direction of motion will tend to assist the force exerted by the person, who will as a result need to exert less force than would otherwise be required. In every case the net force, which is the sum of all the forces acting on the object, must be 20 Newtons, which is the product of its mass and its acceleration. The other forces might act in the direction of the acceleration or in the direction opposite the acceleration; in every case person pulling on the object must exert exactly enough force that the net force will be 20 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The force does not need to necessarily exceed 20 Newtons, just meet the requirement of 20 Newtons to begin to move the object. ------------------------------------------------ Self-critique rating: 2
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Given Solution: Since the 10 Newton frictional force is in the direction opposite to motion, and since the acceleration is in the same direction as the motion, the frictional force is opposed to the accelerating force. If the direction of motion is taken as positive, then the frictional force will be in the negative direction and can be denoted fFrict = - 10 Newtons. To achieve the given acceleration the net force on the object must be net force = 10 kg * (+2 m/s^2) = +20 kg * m/s^2 = +20 Newtons. In order to achieve the +20 Newton net force when there is already a frictional force of -10 Newtons, it should be clear that a force of +30 Newtons is required. This result can be interpreted as follows: The person must exert 10 Newtons of force to overcome friction and another 20 Newtons to achieve the required net force. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I should explain my math better to show how I determined my answers. ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: `q004. How can we write an equation to solve this problem? Hint: What equation would relate the net force Fnet, the force F exerted by the person and the force fFrict of friction? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet = F + fFrict Net Force = Force + Friction Force confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If Fnet is the net force and F the force actually exerted by the person, then Fnet = F + fFrict. That is, the net force is the sum of the force exerted by the person and the frictional force. We know that Fnet is +20 Newtons and fFrict is -10 Newtons, so we have the equation 20 Newtons = F + (-10 Newtons). Solving for F we see that F = 20 Newtons + 10 Newtons = 30 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. If a constant net force of 12 Newtons acts on a cart of mass 6 kg, then at what rate does the velocity of the cart change? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet = m x a Acceleration = Fnet / m A = 12 N / 6 kg A = 2 m/s/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The velocity of the cart will change at a rate a which is related to the net force and the mass by Fnet = m * a. Thus a = Fnet / m = 12 Newtons / (6 kg) = 12 kg * m/s^2 / (6 kg) = 2 m/s^2. We note that the force unit Newtons is broken down to its fundamental units of kg * m / s^2 in order to perform the unit calculation. Dividing kg * m / s^2 by kg we have (kg / kg) * m/s^2 = m/s^2. It is important to always do the unit calculations. This habit avoids a large number of errors and also can be used to reinforce our understanding of the relationships in a problem or situation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. If a force of 50 Newtons is exerted in the direction of the object's motion by a person, on a 20 kg object, and if friction exerts a force of 10 Newtons opposed to the direction of motion, then what will be the acceleration of the object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet = F + fFrict 50 Newtons + (-10 Newtons) = 40 Newtons F = mass x acceration A = F / m 40 Newtons / 20 kg = 2 m/s/s A = 2 m/s/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The object will accelerate at a rate determined by Newton's Second Law, Fnet = m * a. The acceleration will therefore be a = Fnet / m. The net force on the object will be the sum of the 50 Newton force in the direction of motion and the 10 Newton force opposed to the direction of motion. If we take the direction of motion as positive, then the net force is Fnet = 50 N - 10 N = 40 N. It follows that the acceleration is a = Fnet / m = 40 N / (20 kg) = 40 kg m/s^2 / (20 kg) = 2 m/s^2. STUDENT COMMENT: Woops. I added the friction instead of subtracting. So if friction is acting on the object then we subtract it from the force on the object in the direction of motion? I guess it makes since. ... 'sense', not 'since' (I don't usually comment on grammar or incorrect words but I see this one a lot) INSTRUCTOR RESPONSE If we take the direction of motion as positive, then the force in the direction of motion is positive and the frictional force, which acts in the direction opposite motion, is negative. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. If a force of 50 Newtons is exerted opposite to the direction of the object's motion by a person, on a 20 kg object, and if friction exerts a force of 10 Newtons opposed to the direction of motion, then what will be the acceleration of the object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet = F + fFrict -50 Newtons + -10 Newtons = - 60 Newtons The Net force is negative because the forces being applied are in opposition the direction the object is moving, which we assume to be positive. F = m x a A = F / m A = - 60 N / 20 kg A = -3 m/s/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we take the direction of motion to be positive, then since both the 50 Newton force and the 10 Newton force are opposed to the direction of motion the net force must be net force = -50 Newtons - 10 Newtons = -60 Newtons. The acceleration of the object will therefore be a = Fnet / m = -60 Newtons / (10 kg) = -60 kg * m/s^2 / (20 kg) = -3 m/s^2. The fact that the acceleration is opposed to the direction of motion indicates that the object will be slowing down. The force exerted by the person, being in the direction opposite to that of the motion, is seen to be a retarding force, as is friction. So in this case the person is aided by friction in her apparent goal of stopping or at least slowing the object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. If a 40 kg object is moving at 20 m/s, then how long will a take a net force of 20 Newtons directed opposite to the motion of the object to bring the object to rest? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since the net force is acting against the movement of the object, we write net force as -20 Newtions. Fnet = m x a A = F / m A = -20 N / 40 kg A = -0.5 m/s/s Now that we have the acceleration we can use the uniformly accelerated motion equation to determine the time it will take the object to come to rest. Vf= v0 + a 'dt Soving for 'dt, we get the equation: 'dt = (vf - v0) / a v0 = 20 m/s vf = 0 m/s (Seeing as the object will be at rest.) a = -0.5 m/s/s 'dt = (0 m/s - 20 m/s) / -0.5 m/s/s -20 m/s / -0.5 m/s/s = 40 seconds 'dt = 40 seconds for the object to come to rest. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The force on the object is in the direction opposite its motion, so if the direction of motion is taken to be positive the force is in the negative direction. We therefore write the net force as Fnet = -20 Newtons. The acceleration of the object is therefore a = Fnet / m = -20 Newtons / 40 kg = -20 kg * m/s^2 / (40 kg) = -.5 m/s^2. We can therefore describe uniformly accelerated motion of the object as v0 = 20 m/s, vf = 0 (the object comes to rest, which means its velocity ends up at 0), a = -.5 m/s^2. We can then reason out the time required from the -20 m/s change in velocity and the -.5 m/s^2 acceleration, obtaining `dt = 40 seconds. We can confirm our reasoning using the equation vf = v0 + a `dt: Solving for `dt we obtain `dt = (vf - v0) / a = (0 m/s - 20 m/s) / (-.5 m/s^2) = -20 m/s / (-.5 m/s^2) = 40 m/s * s^2 / m = 40 s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q009. If we wish to bring an object with mass 50 kg from velocity 10 m/s to velocity 40 m/s in 5 seconds, what constant net force would be required? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet = mass x acceleration Acceleration = change in velocity / time interval A = (40 m/s - 10 m/s) / 5 seconds A = 6 m/s/s Fnet = 50 kg x 6 m/s/s Fnet = 300 kgm/s/s = 300 Newtons confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The net force would be Fnet = m * a. The acceleration of the object would be the rate which its velocity changes. From 10 m/s to 40 m/s the change in velocity is +30 m/s; to accomplish this in 5 seconds requires average acceleration 30 m/s / (5 s) = 6 m/s^2. Thus the net force required is Fnet = m * a = 50 kg * 6 m/s^2 = 300 kg m/s^2 = 300 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q010. If a constant net force of 50 Newtons brings an object to rest in four seconds from an initial velocity of 8 meters/second, then what is the mass of the object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Acceleration = change in velocity / time interval A = ( 0 m/s - 8 m/s) / 4 sec A = -2 m/s/s F = m x a m = F / a m = -50 N / -2 m/s/s m = 25 kg confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We know the net force and we have the information required to calculate the acceleration. We will therefore be able to find the mass using Newton's Second Law Fnet = m * a. We first find the acceleration. The change in velocity from 8 m/s to rest is -8 m/s, and this occurs in 4 seconds. The acceleration is therefore -8 m/s / (4 s) = -2 m/s^2. The 50 Newton net force must be in the same direction as the acceleration, so we have Fnet = -50 Newtons. We obtain the mass by solving Newton's Second Law for m: m = Fnet / a = -50 N / (-2 m/s^2) = -50 kg m/s^2 / (-2 m/s^2) = 25 kg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Questions related to q_a_ 1. If a 12 kg object accelerates at 5 m/s^2 and you are exerting a force of -20 N on it, then what is the sum of the other forces acting on the object? Give a plausible interpretation of this situation. 2. What force does gravity exert on a 50 kg object? If the object is accelerating upward at 12 m/s^2, what must be the total of the nongravitational forces acting on it? Questions related to Introductory Problem Sets 1. If an object of mass 5 Kg and initially at rest is pushed by a net force of 20 Newtons for 7 seconds, what are its acceleration, its final velocity, its average velocity, and the distance it travels? 2. An object, initially at rest, is acted upon by a net force of 15 Newtons. The object has mass 3 kilograms. The force acts for 7 seconds. What velocity will the object attain and how far will it travel during this time? What kinetic energy will it attain? How much work is done on the object by the net force during this interval? Questions related to text Questions/problems for Principles of Physics Students 1. A bee flies at 10 km / hr. How long does it take to fly 12 meters from its hive to your hat? 2. Convert 35 mi / h to m / s, to km / hr and to ft / s. 3. Between clock times t_1 = 5.0 sec and t_2 = 7.8 sec, a ball travels from position x_1 = 28 cm to position x_2 = -12 cm. What is its average velocity during this interval? Can you determine its average speed from this information? 4. A dragster accelerates from rest to 150 km / hr is 4.2 s. What is its average acceleration in km / hr^2, in m/s^2, and in ft / s^2? 5. A car slows from 25 m/s to rest while traveling 100 meters, accelerating at a constant velocity. What was its acceleration? 6. A car speeds up from rest to 95 km / hr in 6.2 s. What is its average acceleration? Questions/problems for General College Physics Students 7. Two trains, initially 12.4 km apart, approach one another on parallel tracks. Each is moving at 80 km / s relative to the ground. How long will it be before they reach one another? 8. A pickup truck moving at 60 km / hr strikes a tree, bringing the passenger compartment (and the passenger) to rest in a distance of .50 meters. What is the average acceleration of the driver? What is this acceleration in 'g's', where a 'g' is 9.8 m/s^2? Questions/problems for University Physics Students 9. You ride the first 10 miles of a 20-mile ride at average speed 8 m/h. What must be your average speed on the last 10 miles in order to average 10 m/h for the entire trip? What is the greatest average speed you could possibly attain, given these conditions? 10. A train moving at 25 m/s is 200 meters behind a train which is moving at 15 m/s, when the first train hits its brakes. The first train accelerates at -.100 m/s^2 while the second train continues moving at constant velocity. Will there be a collision? If so, where will it take place? Describe a single graph that depicts the position vs. clock time of the front of the first train and the back of the second.. 11. An automobile and a truck are traveling in the same direction on two lanes of a highway, both moving at 25 m/s. The automobile is behind the truck. They both hit their brakes at the same instant. The magnitude of the truck's acceleration is 3 m/s^2, the magnitude of the car's acceleration is 2 m/s^2. By the time the truck has moved 50 meters, the car has caught up. How far behind the truck was the car? How long did it take for the car to catch the truck? How fast is each moving at the instant the car overtakes the truck? Describe the position vs. clock time graph which depicts the motion of both vehicles. Questions related to key systems 1. If a ball accelerates from rest through a distance of 30 cm while a 'pearl pendulum' of length 8 cm, released simultaneously with the ball, strikes the bracket 7 times, what is the average acceleration of the ball? 2. (note that the problems on proportionality in the document at h ttp://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/lib1/lib1_qa29.htm might be useful in understanding the concept of proportionality and variation). Suppose we conduct the following experiment: On a fixed ramp we release an object from rest. We determine how long it takes the object to travel various distances down this incline. From this information we calculate the final velocities attained for various distances, assuming that the acceleration in each case is uniform. We graph final velocity vs. distance, and find that the graph is clearly not a straight line. We then graph the square of the final velocity vs. distance and find that the graph is a straight line. We use our data to test two hypotheses: Hypothesis 1: The change in the velocity of a uniformly accelerating object is proportional to the product of the acceleration and the time interval. Hypothesis 2: The change in the squared velocity of a uniformly accelerating object is proportional to the product of the acceleration and the time interval. Do these results support Hypothesis 1 but not Hypothesis 2, Hypothesis 2 but no Hypothesis 1, both Hypotheses 1 and 2, or neither of the two hypotheses? Explain. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Questions related to q_a_ 1. If a 12 kg object accelerates at 5 m/s^2 and you are exerting a force of -20 N on it, then what is the sum of the other forces acting on the object? Give a plausible interpretation of this situation. 2. What force does gravity exert on a 50 kg object? If the object is accelerating upward at 12 m/s^2, what must be the total of the nongravitational forces acting on it? Questions related to Introductory Problem Sets 1. If an object of mass 5 Kg and initially at rest is pushed by a net force of 20 Newtons for 7 seconds, what are its acceleration, its final velocity, its average velocity, and the distance it travels? 2. An object, initially at rest, is acted upon by a net force of 15 Newtons. The object has mass 3 kilograms. The force acts for 7 seconds. What velocity will the object attain and how far will it travel during this time? What kinetic energy will it attain? How much work is done on the object by the net force during this interval? Questions related to text Questions/problems for Principles of Physics Students 1. A bee flies at 10 km / hr. How long does it take to fly 12 meters from its hive to your hat? 2. Convert 35 mi / h to m / s, to km / hr and to ft / s. 3. Between clock times t_1 = 5.0 sec and t_2 = 7.8 sec, a ball travels from position x_1 = 28 cm to position x_2 = -12 cm. What is its average velocity during this interval? Can you determine its average speed from this information? 4. A dragster accelerates from rest to 150 km / hr is 4.2 s. What is its average acceleration in km / hr^2, in m/s^2, and in ft / s^2? 5. A car slows from 25 m/s to rest while traveling 100 meters, accelerating at a constant velocity. What was its acceleration? 6. A car speeds up from rest to 95 km / hr in 6.2 s. What is its average acceleration? Questions/problems for General College Physics Students 7. Two trains, initially 12.4 km apart, approach one another on parallel tracks. Each is moving at 80 km / s relative to the ground. How long will it be before they reach one another? 8. A pickup truck moving at 60 km / hr strikes a tree, bringing the passenger compartment (and the passenger) to rest in a distance of .50 meters. What is the average acceleration of the driver? What is this acceleration in 'g's', where a 'g' is 9.8 m/s^2? Questions/problems for University Physics Students 9. You ride the first 10 miles of a 20-mile ride at average speed 8 m/h. What must be your average speed on the last 10 miles in order to average 10 m/h for the entire trip? What is the greatest average speed you could possibly attain, given these conditions? 10. A train moving at 25 m/s is 200 meters behind a train which is moving at 15 m/s, when the first train hits its brakes. The first train accelerates at -.100 m/s^2 while the second train continues moving at constant velocity. Will there be a collision? If so, where will it take place? Describe a single graph that depicts the position vs. clock time of the front of the first train and the back of the second.. 11. An automobile and a truck are traveling in the same direction on two lanes of a highway, both moving at 25 m/s. The automobile is behind the truck. They both hit their brakes at the same instant. The magnitude of the truck's acceleration is 3 m/s^2, the magnitude of the car's acceleration is 2 m/s^2. By the time the truck has moved 50 meters, the car has caught up. How far behind the truck was the car? How long did it take for the car to catch the truck? How fast is each moving at the instant the car overtakes the truck? Describe the position vs. clock time graph which depicts the motion of both vehicles. Questions related to key systems 1. If a ball accelerates from rest through a distance of 30 cm while a 'pearl pendulum' of length 8 cm, released simultaneously with the ball, strikes the bracket 7 times, what is the average acceleration of the ball? 2. (note that the problems on proportionality in the document at h ttp://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/lib1/lib1_qa29.htm might be useful in understanding the concept of proportionality and variation). Suppose we conduct the following experiment: On a fixed ramp we release an object from rest. We determine how long it takes the object to travel various distances down this incline. From this information we calculate the final velocities attained for various distances, assuming that the acceleration in each case is uniform. We graph final velocity vs. distance, and find that the graph is clearly not a straight line. We then graph the square of the final velocity vs. distance and find that the graph is a straight line. We use our data to test two hypotheses: Hypothesis 1: The change in the velocity of a uniformly accelerating object is proportional to the product of the acceleration and the time interval. Hypothesis 2: The change in the squared velocity of a uniformly accelerating object is proportional to the product of the acceleration and the time interval. Do these results support Hypothesis 1 but not Hypothesis 2, Hypothesis 2 but no Hypothesis 1, both Hypotheses 1 and 2, or neither of the two hypotheses? Explain. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Phy201+L 010. `query 10
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Given Solution: First way: KE change is equal to the work done by the net force, which is net force * displacement, or Fnet * `ds. Second way: KE change is also equal to Kef - KE0 = .5 m vf^2 - .5 m v0^2. ** STUDENT QUESTION: I wasn’t sure what equation to use to find KE the second way. What does Kef stand for? INSTRUCTOR RESPONSE: In general f stands for 'final' and 0 for 'initial'. Just as v0 and vf stand for initial and final velocities, we'll use KEf and KE0 to stand for initial and final kinetic energies. STUDENT QUESTION: Ok I know the other equation now but I still don’t really understand it. How come you multiply each velocity by 0.5? I don’t really understand the second equation KE = 1/2 m v^2 INSTRUCTOR RESPONSE On one level, KE = 1/2 m v^2 is simply a formula you have to know. It isn't hard to derive that formula, as you'll see soon, and the 1/2 arises naturally enough. A synopsis of the derivation: If force F_net is applied to mass m through displacement `ds then: a = F_net / m, and vf^2 = v0^2 + 2 a `ds It's not difficult to rearrange the result of these two equations to get F_net * `ds = 1/2 m vf^2 - 1/2 m v0^2. You'll see the details soon, but that's where the formula KE = 1/2 m v^2 comes from; the 1/2 or 0.5 is part of the solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q (This question applies primarily to General College Physics students and University Physics students, though Principles of Physics students are encouraged, if they wish, to answer the question). In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vf^2 = v0^2 + 2 a 'ds, so a 'ds = (vf^2 - v0^2) /2, so a 'ds is proportional to the change in v^2. Fnet = m x a, so Fnet x 'ds = m x a 'ds, so Fnet 'ds is proportional to a 'ds. So if Fnet is proportional to a 'ds, it is also proportional to v^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In a nutshell: since vf^2 = v0^2 + 2 a `ds, a `ds = 1/2 (vf^2 - v0^2), so a `ds is proportional to the change in v^2 since F_net = m a, F_net * `ds = m a * `ds so F_net * `ds is proportional to a * `ds Thus F_net `ds is proportional to a * `ds, which in turn is proportional to the change in v^2. Thus F_net `ds is proportional to the change in v^2. More detail: It's very important in physics to be able to think in terms of proportionality. To say that y is proportional to x is to say that for some k, y = k x. That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that for some k, a * `ds = k * ( change in v^2)--i.e., that a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for the specific k value k = 1/2. Now since Fnet = m a, we conclude that Fnet * `ds = m a * `ds and since a `ds = k * ( change in v^2) for the specific k value k = 1/2, we substitute for a * `ds to get Fnet `ds = m * k * (change in v^2), for k = 1/2. Now m and k are constants, so m * k is constant. We can therefore revise our value of k, so that it becomes m * 1/2 or m / 2 With this revised value of k we have Fnet * `ds = k * (change in v^2), where now k has the value m / 2. That is, we don't expect Fnet * `ds to be proportional to the change in velocity v, but to the change in the square v^2 of the velocity. STUDENT COMMENT: I am still a bit confused. Going through the entire process I see how these values correlate but on my own I am not coming up with the correct solution. I am getting lost after we discover the a `ds is a constant multiple of (1/2) the change in v^2. Is it that I should simply substitute the k into the equation? Or am I missing something else? INSTRUCTOR RESPONSE: The short answer is that by the fourth equation of uniformly accelerated motion, a `ds = 1/2 (vf^2 - v0^2), which is half the change in v^2, so that a `ds is proportional to the change in v^2. (The proportionality constant between a `ds and change in v^2 is the constant number 1/2). F_net = m a, where m is the mass of the object. So F_net is proportional to a. (The proportionality constant between F_net and a is the constant mass m). Thus F_net `ds is proportional to a `ds, which we have seen is proportional to the change in v^2. The conclusion is the F_net `ds is proportional to the change in v^2. (The proportionality constant between F_net `ds and change in v^2 is 1/2 m.) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ------------------------------------------------ Self-critique Rating: ********************************************* Question: How do our experimental results confirm or cause us to reject this hypothesis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since all of the objects started from rest, vf - v0 (v0 will be zero as all of the objects started from rest) then vf^2 = v^2. This would show us that a'ds is proportional to v^2. This gives us a linear graph. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The explanation for this result: On a ramp with fixed slope the acceleration is constant so a `ds is simply proportional to `ds specifically a `ds = k * `ds for k = a. In the preceding question we saw why a * `ds = k * (change in v^2), with k = 1/2. In our experiment the object always accelerated from rest. So the change in v^2 for each trial would be from 0 to vf^2. the change would therefore be just change in v^2 = vf^2 - 0^2 = vf^2. Thus if a `ds is proportional to the change in vf^2, our graph of vf^2 vs. a `ds should be linear. The slope of this graph would just be our value of k in the proportionality a * `ds = k * (change in v^2), where as we have seen k = 1/2 We wouldn't even need to determine the actual value of the acceleration a. To confirm the hypothesis all we need is a linear graph of vf^2 vs. `ds. (we could of course use that slope with our proportionality to determine a, if desired) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I need to be more through, but I did not understand this question very well. ------------------------------------------------ Self-critique Rating:2 ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGeneral College Physics and Principles of Physics: convert 35 mi/hr to km/hr, m/s and ft/s. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 mi = 1.609 km 35 mi/hr x 1.609 = 56.315 km /hr 1 km = 1000 meters 56.315 km/hr x 1000 m = 56315 m /hr 1 hr = 3600 seconds 56315 m / hr / 3600 sec = 15.64 m/s 1 m = 3.28 feet 15.64 m/s x 3.28 ft = 51.30 ft/sec confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe need a conversions between miles and meters, km and ft, and we also need conversions between hours and seconds. We know that 1 mile is 5280 ft, and 1 hour is 3600 seconds. We also know that 1 inch is 2.54 cm, and of course 1 foot is 12 inches. 1 mile is therefore 1 mile * 5280 ft / mile = 5280 ft, 5280 ft = 5280 ft * 12 in/ft * 2.54 cm / in = 160934 cm, which is the same as 160934 cm * 1 m / (100 cm) = 1609.34 m, which in turn is the same as 1609.34 m * 1 km / (1000 m) = 1.60934 km. Thus 35 mi / hr = 35 mi / hr * (1.60934 km / 1 mi) = 56 (mi * km / (mi * hr) ) = 56 (mi / mi) * (km / hr) = 56 km / hr. We can in turn convert this result to m / s: 56 km/hr * (1000 m / km) * (1 hr / 3600 sec) = 15.6 (km * m * hr) / (hr * km * sec) = 15.6 (km / km) * (hr / hr) * (m / s) = 15.6 m/s. The original 35 mi/hr can be converted directly to ft / sec: 35 mi/hr * ( 5280 ft / mi) * ( 1 hr / 3600 sec) = 53.33 ft/sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGen phy and prin phy prob 2.16: sports car rest to 95 km/h in 6.2 s; find acceleration YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 95 km /hr 95 x 1000 = 95000 meters 95000 m / 3600 sec = 26.4 m/s v0 = 0 m/s vf = 26.4 m/s 26.4 m/s / 6.2 sec = 4.26 m/s/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 95 km/hr = 95,000 m / (3600 sec) = 26.3 m/s. So change in velocity is `dv = 26.3 m/s = 0 m/s = 26.3 m/s. Average acceleration is aAve = `dv / `dt = 26.3 m/s / (6.2 s) = 4.2 m/s^2. Extension: One 'g' is the acceleration of gravity, 9.8 m/s^2. So the given acceleration is -4.2m/s^2 / [ (9.8 m/s^2) / 'g' ] = -.43 'g'. STUDENT QUESTION: How did we know that the final velocity was 0? INSTRUCTOR RESPONSE: The final velocity was 0 because the car came to rest. Summary of what we were given: Initial velocity is 95 km/hr, or 26.3 m/s. Final velocity is 0, since the car came to rest. The velocity makes this change in a time interval of 6.2 seconds. We can easily reason out the result using the definition of acceleration: The acceleration is the rate at which velocity changes with respect to clock time, which by the definition of rate is (change in velocity) / (change in clock time) The change in velocity from the initial 0 m/s to the final 26.3 m/s is 26.3 m/s, so acceleration = change in velocity / change in clock time = 26.3 m/s / (6.2 s) = 4.2 m/s^2. We could also have used the equations of uniformly accelerated motion, with vf = 26.3 m/s, v0 = 0 and `dt = 6.2 seconds. However in this case it is important to understand that the definition of acceleration can be applied directly, with no need of the equations. (solution using equations: 2d equation is vf = v0 + a `dt, which includes our three known quantities; solving for a we get a = (vf - v0) / `dt = (26.3 m/s - 0 m/s) / (6.2 s) = 4.2 m/s^2.) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: univ phy 2.66 train 25m/s 200 m behind 15 m/s train, accel at -.1 m/s^2. Will the trains collide and if so where? Describe your graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we assume the passenger train is at position x = 0 at clock time t = 0 we conclude that the position function is x(t) = x0 + v0 t + .5 a t^2; in this case a = -.1 m/s&2 and x0 was chosen to be 0 so we have x(t) = 25 m/s * t + .5 * (-.1m/s^2) * t^2 = 25 m/s * t - .05 m/s^2 * t^2. To distinguish the two trains we'll rename this function x1(t) so that x1(t) = 25 m/s * t - .05 m/s^2 * t^2. At t = 0 the freight train, which does not change speed so has acceleration 0 and constant velocity 15 m/s, is 200 m ahead of the passenger train, so the position function for the freight train is x2(t) = 200 m + 15 m/s * t . The positions will be equal if x1 = x2, which will occur at any clock time t which solves the equation 25 t - .05 t^2 = 200 + 15 t(units are suppressed here but we see from the units of the original functions that solutions t will be in seconds). Rearranging the equation we have -.05 t^2 + 10 t - 200 = 0. The quadratic formula tells us that solutions are t = [ - 10 +- sqrt( 10^2 - 4 * (-.05) * (-200) ) ] / ( 2 * .05 ) Simplifying we get solutions t = 22.54 and t = 177.46. At t = 22.54 seconds the trains will collide. Had the trains been traveling on parallel tracks this would be the instant at which the first train overtakes the second. t = 177.46 sec would be the instant at which the second train again pulled ahead of the slowing first train. However since the trains are on the same track, the accelerations of both trains will presumably change at the instant of collision and the t = 177.46 sec solution will not apply. GOOD STUDENT SOLUTION: for the two trains to colide, the 25 m/s train must have a greater velocity than the 15 m/s train. So I can use Vf = V0 + a('dt). 15 = 25 + (-.1)('dt) -10 = -.('dt) 'dt = 100 so unless the displacement of the 25 m/s train is greater than the 15 m/s train in 100 s, their will be no colision. 'ds = 15 m/s(100) + 200 m 'ds = 1700 m 'ds = 25 m/s(100) + .5(-.1)(100^2) = 2000 m. The trains collide. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: