#$&* course Phy 201 10/4 at 4:00 010. Note that there are 10 questions in this set.
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Given Solution: The net force on the block is the product F = m * a of its 10 kg mass and its 2 m/s^2 acceleration. The net force is therefore • F = 10 kg * 2 m/s^2 = 20 kg * m / s^2. The unit of force, which is the product of a quantity in kg and another quantity in m/s^2, is just the algebraic product kg * m/s^2 of these two units. This unit, the kg * m / s^2, is called a Newton. So the net force is 20 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*:K ********************************************* 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: We would use the same formula as the last problem and get 20 Newtons again. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: This depends on what forces might be resisting the acceleration of the object. • If the object is accelerating on a surface of some type, then there is a good chance that a frictional force is opposing the motion. In this case the person would have to exert more force than if friction was not present. • If the object is being pulled upward against the force of gravity, then force must be sufficient to counteract the gravitational force, and in addition to accelerate the object in the upward direction. • If the object is being pulled downhill, the force exerted by gravity has a component in the direction of motion. The component of the gravitational force in the direction of motion will tend to assist the force exerted by the person, who will as a result need to exert less force than would otherwise be required. In every case the net force, which is the sum of all the forces acting on the object, must be 20 Newtons, which is the product of its mass and its acceleration. The other forces might act in the direction of the acceleration or in the direction opposite the acceleration; in every case person pulling on the object must exert exactly enough force that the net force will be 20 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see now that with the given information we cannot tell exactly how much force the person must exert but it needs to be enough for the net force to equal 20 Newtons. ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q003. If friction exerts a force of 10 Newtons in the direction opposite the motion of a 10 kg object, then how much force must be exerted by someone pulling on it to accelerate the 10 kg object at 2 m/s^2, with the acceleration in the same direction as the motion? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since the required net force is 20 newtons, which we found out from the last two problems, we know that the total net force must equal this. Therefore if the frictional force is -10 Newtons or 10 newtons in the opposite direction, we will need the person to exert a total of 30 Newtons. (10 N to make up for the force pushing in the opposite direction and 20 for the net force). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Since the 10 Newton frictional force is in the direction opposite to motion, and since the acceleration is in the same direction as the motion, the frictional force is opposed to the accelerating force. If the direction of motion is taken as positive, then the frictional force will be in the negative direction and can be denoted • fFrict = - 10 Newtons. To achieve the given acceleration the net force on the object must be • net force = 10 kg * (+2 m/s^2) = +20 kg * m/s^2 = +20 Newtons. In order to achieve the +20 Newton net force when there is already a frictional force of -10 Newtons, it should be clear that a force of +30 Newtons is required. This result can be interpreted as follows: The person must exert 10 Newtons of force to overcome friction and another 20 Newtons to achieve the required net force. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*:K ********************************************* 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 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: If Fnet is the net force and F the force actually exerted by the person, then • Fnet = F + fFrict. That is, the net force is the sum of the force exerted by the person and the frictional force. We know that Fnet is +20 Newtons and fFrict is -10 Newtons, so we have the equation • 20 Newtons = F + (-10 Newtons). Solving for F we see that F = 20 Newtons + 10 Newtons = 30 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*:K ********************************************* 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: Since acceleration is the average rate of velocity change, we will use the formula a= F/m to get: 12/6= 2m/s/s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The velocity of the cart will change at a rate a which is related to the net force and the mass by Fnet = m * a. Thus a = Fnet / m = 12 Newtons / (6 kg) = 12 kg * m/s^2 / (6 kg) = 2 m/s^2. We note that the force unit Newtons is broken down to its fundamental units of kg * m / s^2 in order to perform the unit calculation. Dividing kg * m / s^2 by kg we have (kg / kg) * m/s^2 = m/s^2. It is important to always do the unit calculations. This habit avoids a large number of errors and also can be used to reinforce our understanding of the relationships in a problem or situation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*:K ********************************************* 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: We first need to figure out the Fnet which would be 50N-10N= 40 Newtons. Then we will use the formula a=F/m to get 40N/20kg= 2 m/s/s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: The object will accelerate at a rate determined by Newton's Second Law, Fnet = m * a. The acceleration will therefore be a = Fnet / m. The net force on the object will be the sum of the 50 Newton force in the direction of motion and the 10 Newton force opposed to the direction of motion. If we take the direction of motion as positive, then the net force is Fnet = 50 N - 10 N = 40 N. It follows that the acceleration is a = Fnet / m = 40 N / (20 kg) = 40 kg m/s^2 / (20 kg) = 2 m/s^2. STUDENT COMMENT: Woops. I added the friction instead of subtracting. So if friction is acting on the object then we subtract it from the force on the object in the direction of motion? I guess it makes since. ... 'sense', not 'since' (I don't usually comment on grammar or incorrect words but I see this one a lot) INSTRUCTOR RESPONSE If we take the direction of motion as positive, then the force in the direction of motion is positive and the frictional force, which acts in the direction opposite motion, is negative. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*:K ********************************************* 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: We first need to find the Fnet which would be 50+10= -60 N. It is -60 because the force is being exerted in the opposite direction. We then use a= F/m to get: -60/ 20= -3m/s/s for acceleration. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: If we take the direction of motion to be positive, then since both the 50 Newton force and the 10 Newton force are opposed to the direction of motion the net force must be net force = -50 Newtons - 10 Newtons = -60 Newtons. The acceleration of the object will therefore be a = Fnet / m = -60 Newtons / (10 kg) = -60 kg * m/s^2 / (20 kg) = -3 m/s^2. The fact that the acceleration is opposed to the direction of motion indicates that the object will be slowing down. The force exerted by the person, being in the direction opposite to that of the motion, is seen to be a retarding force, as is friction. So in this case the person is aided by friction in her apparent goal of stopping or at least slowing the object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*:K ********************************************* 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: m= 40 kg Fnet= -20 N v0= 20m/s vf=0m/s a= F/m= -20/40= -.5m/s/s We now need to find ‘dt. We can do this by using the v0=20m/s, vf=0m/s and a= -.5m/s/s. The equation is: a= (vf-v0)/ ‘dt= -.5m/s/s= (0-20)/ ‘dt - - - -.5= -20/ ‘dt --- -.5(‘dt)=-20 – ‘dt= -20/-.5 - - 40 seconds. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: The force on the object is in the direction opposite its motion, so if the direction of motion is taken to be positive the force is in the negative direction. We therefore write the net force as Fnet = -20 Newtons. The acceleration of the object is therefore a = Fnet / m = -20 Newtons / 40 kg = -20 kg * m/s^2 / (40 kg) = -.5 m/s^2. We can therefore describe uniformly accelerated motion of the object as v0 = 20 m/s, vf = 0 (the object comes to rest, which means its velocity ends up at 0), a = -.5 m/s^2. We can then reason out the time required from the -20 m/s change in velocity and the -.5 m/s^2 acceleration, obtaining `dt = 40 seconds. We can confirm our reasoning using the equation vf = v0 + a `dt: Solving for `dt we obtain `dt = (vf - v0) / a = (0 m/s - 20 m/s) / (-.5 m/s^2) = -20 m/s / (-.5 m/s^2) = 40 m/s * s^2 / m = 40 s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*:K ********************************************* 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: We need to first find the acceleration. A= (vf-v0)/ ‘dt= (40m/s-10m/s)/5= 30m/s /5= 6 m/s/s We then use the formula F= ma- - F= 50kg (6m/s/s)= 300 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The net force would be Fnet = m * a. The acceleration of the object would be the rate which its velocity changes. From 10 m/s to 40 m/s the change in velocity is +30 m/s; to accomplish this in 5 seconds requires average acceleration 30 m/s / (5 s) = 6 m/s^2. Thus the net force required is Fnet = m * a = 50 kg * 6 m/s^2 = 300 kg m/s^2 = 300 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*:K ********************************************* 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: We must first figure out the acceleration: a= (vf-v0)/’dt - - (8-0)/ 4= 2m/s/s We will then use the formula: a= F/m- - 2m/s/s= 50/m - - 2m= 50 - - m= 25 kg confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: We know the net force and we have the information required to calculate the acceleration. We will therefore be able to find the mass using Newton's Second Law Fnet = m * a. We first find the acceleration. The change in velocity from 8 m/s to rest is -8 m/s, and this occurs in 4 seconds. The acceleration is therefore -8 m/s / (4 s) = -2 m/s^2. The 50 Newton net force must be in the same direction as the acceleration, so we have Fnet = -50 Newtons. We obtain the mass by solving Newton's Second Law for m: m = Fnet / a = -50 N / (-2 m/s^2) = -50 kg m/s^2 / (-2 m/s^2) = 25 kg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: OK 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. We would use the formula: F= m a to get 60 N. So the force acting on the object would be 80 to makeup for the negative force I am exerting. 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? F= 50 * 12= 600 N Gravity would emit 490 N (9.8*50). Therefore the total of the nongravitational forces would be 110 N. 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? We would first find the acceleration by using the formula a= F/m. We then divide 20 N/5kg = 4m/s/s. We then use this to find the final velocity. A= (vf-v0)/ ‘dt - -4m/s/s= (vf-0)/7 - -28m/s= vf. We can now find the average velocity: (vf+v0)/2 - - (28+0)/2= 16m/s Now we will find the distance traveled by using: vAve= ‘ds/ ‘dt- - - 16m/s/s= ‘ds/ 7 =112m = ‘ds 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? First I will find acceleration: a= F/m, so 15/3 equals 15m/s/s. We will use this to find the vf. 15m/s/s= (vf-0)/7 - - 105m/s=vf We first have to find the vAve. 105+0/2= 52.5 m/s Now we can find the ‘ds. 52.5= ‘ds/ 7s - - 167.5 m= ‘ds • What kinetic energy will it attain? We can use the formula: KE= ½ m v0^2. KE= ½(3)(0)= 1.5 • How much work is done on the object by the net force during this interval? ‘dw= F ‘ds, so we would multiply 15 N*167.5= 2,512.5 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? We first have to divide 12.4 by 2 to get 6.2 which is the distance until they reach one another. We will use the formula: ‘dt= sqrt 2(‘ds)/a - - ‘dt= sqrt 2(6.2)/80 = sqrt 12.4/80 = sqrt .155. The trains will reach each other in .4 seconds. 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? .50 m= 500 km We will then find ‘dt by dividing 60 km/hr by .50 meters. This equals .12 hr. We will then use: ‘dv/ ‘dt to find the acceleration- 60/.12= 500km/hr/hr. The acceleration in ‘g’s’ would be 9.8’s’ or times .12 which would give us: 1.176km/hr/hr 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? I am unsure how to do this problem completely. Please advice. 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. I would say that it supports Hypothesis 2, but not 1 because the formula a= ‘dv/’dt would support this and the graph. "