#$&* course Phy 121 6/24 around 2:25. ?????I assumed the bottom questions were like those included in the previous qa assignments but if you would like me to submit those as well I will.???? 010. Note that there are 12 questions in this set.
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Given Solution: The net force on the block is the product F = m * a of its 10 kg mass and its 2 m/s^2 acceleration. The net force is therefore • F = 10 kg * 2 m/s^2 = 20 kg * m / s^2. The unit of force, which is the product of a quantity in kg and another quantity in m/s^2, is just the algebraic product kg * m/s^2 of these two units. This unit, the kg * m / s^2, is called a Newton. So the net force is 20 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. How much force must be exerted by someone pulling on it to accelerate a 10 kg object at 2 m/s^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The net force will be 20 Newtons because F = m*a, so 10kg * 2m/s^2 = 20kg m/s ^2=20 Newtons. We do not know how much force must be exerted by the person pulling because we do not know how much force is resisting the movement. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique rating: OK ********************************************* 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: The person must exert 30 Newtons because the net force of this object is 20 Newtons. The friction causes the person too have to exert 10 more Newtons to overcome the friction and still have a net force of 20, so 20+10=30 Newtons. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since the 10 Newton frictional force is in the direction opposite to motion, and since the acceleration is in the same direction as the motion, the frictional force is opposed to the accelerating force. If the direction of motion is taken as positive, then the frictional force will be in the negative direction and can be denoted • fFrict = - 10 Newtons. To achieve the given acceleration the net force on the object must be • net force = 10 kg * (+2 m/s^2) = +20 kg * m/s^2 = +20 Newtons. In order to achieve the +20 Newton net force when there is already a frictional force of -10 Newtons, it should be clear that a force of +30 Newtons is required. This result can be interpreted as follows: The person must exert 10 Newtons of force to overcome friction and another 20 Newtons to achieve the required net force. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q004. How can we write an equation to solve this problem? Hint: What equation would relate the net force Fnet, the force F exerted by the person and the force fFrict of friction? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet = F + fFrict. This is because the net force is all the forces on an object. In the previous problem friction was -10 Newtons because it was going in the opposite direction of the acceleration and the person exerted 30 Newtons. This resulted in a net force of 20 Newtons because -10 + 30 = 20 Newtons. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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: The velocity changes at a rate of 2m/s^2 because F = m*a and 12 kg * m/s^2 = 6 kg* a, so 12/6 = 2m/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The velocity of the cart will change at a rate a which is related to the net force and the mass by Fnet = m * a. Thus a = Fnet / m = 12 Newtons / (6 kg) = 12 kg * m/s^2 / (6 kg) = 2 m/s^2. We note that the force unit Newtons is broken down to its fundamental units of kg * m / s^2 in order to perform the unit calculation. Dividing kg * m / s^2 by kg we have (kg / kg) * m/s^2 = m/s^2. It is important to always do the unit calculations. This habit avoids a large number of errors and also can be used to reinforce our understanding of the relationships in a problem or situation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q006. If a force of 50 Newtons is exerted in the direction of the object's motion by a person, on a 20 kg object, and if friction exerts a force of 10 Newtons opposed to the direction of motion, then what will be the acceleration of the object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The acceleration of the given object will be 2m/s^2 because the net force on the object is 40 Newtons. This is because the friction exerts -10 Newtons and the person exerts +50 Newtons, so -10 + 50 = 40 Newtons. Given this fact we cann then plug the information into the following equation: F = m * a 40 Newtons = 20 kg * a 40 kg * m/s^2 = 20 k * a 2m/s^2 = a confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique rating: OK ********************************************* 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: Given that the friction and the motion by the person are goin in the same direction we will then add them, so -50 + -10 = -60 Newtons. The Net force is negative because it is going against the acceleration. Given this fact we can plug the information into the following equation: F = m * a -60 N = 20 kg * a -60 kg * m/s^2 = 20kg * a -3m/s^2 = a The acceleration of the given object will be -3m/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we take the direction of motion to be positive, then since both the 50 Newton force and the 10 Newton force are opposed to the direction of motion the net force must be net force = -50 Newtons - 10 Newtons = -60 Newtons. The acceleration of the object will therefore be a = Fnet / m = -60 Newtons / (10 kg) = -60 kg * m/s^2 / (20 kg) = -3 m/s^2. The fact that the acceleration is opposed to the direction of motion indicates that the object will be slowing down. The force exerted by the person, being in the direction opposite to that of the motion, is seen to be a retarding force, as is friction. So in this case the person is aided by friction in her apparent goal of stopping or at least slowing the object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q008. If a 40 kg object is moving at 20 m/s, then how long will a take a net force of 20 Newtons directed opposite to the motion of the object to bring the object to rest? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given the fact that a = Fnet / m. We can find the acceleration by using this equation. a = Fnet/m a = 20N/40kg a = 20kg*m/s^2 / 40kg a = .5m/s^2 The change in velocity of the object will be -20m/s because its final velocity will be 0 since it is coming to a complete stop. Therefore with a change in velocity of -20m/s and an acceleration of -.5m/s^2, the object will take 40 seconds because -20m/s * -.5m/s^2 = 40 seconds. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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: The acceleration of the given object would be 6m/s^2 because the change in velocity is 30m/s and traveling at this rate for 5 seconds would be 30m/s / 5s = 6m/s^2. We now have the acceleration and the mass of the object we can find the net force using the following equation: F = m*a F = 50kg * 6m/s^2 F = 300kg * m/s^2 F = 300 Newtons. The net force of the given object is 300 Newtons. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The net force would be Fnet = m * a. The acceleration of the object would be the rate which its velocity changes. From 10 m/s to 40 m/s the change in velocity is +30 m/s; to accomplish this in 5 seconds requires average acceleration 30 m/s / (5 s) = 6 m/s^2. Thus the net force required is Fnet = m * a = 50 kg * 6 m/s^2 = 300 kg m/s^2 = 300 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q010. If a constant net force of 50 Newtons brings an object to rest in four seconds from an initial velocity of 8 meters/second, then what is the mass of the object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The acceleration of the given object would be -2m/s^2 because the change in acceleration is -8m/s and the time is 4 seconds, so -8m/s / 4s = -2m/s^2. The net force must also be negative because it is going in the same direction as the acceleration. Given this fact we can use the following equation to find the mass of the given object: F = m*a -50Newtons = m * -2m/s^2 -50 kg * m/s^2 = m * -2m/s^2 25 kg = m The mass of the given object is 25 kg. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating:OK If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q011. A system of mass 16 kg is subjected to opposing forces of 8 Newtons and 4 Newtons. What is its acceleration? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given that both forces (8N and 4N) are opposing the object, then the net force of the object will be -12 Newtons because -8N + -4N = -12N. Given this information we can now use to following equation to find the acceleration: F = m*a -12N = 16kg * a -12 kg * m/s^2 = 16kg * a -.75m/s^2 = a The acceleration of the given object is -.75m/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q012. A 4000 kg mass is to accelerated uniformly from rest to 10 meters/second during a 5-second interval. It experiences an opposing frictional force of 5000 Newtons. How much force must be applied? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The acceleration of the given object is 2m/s^2 because the change in velocity is 10m/s and this happens in 5 seconds, so 10m/s / 5s = 2m/s^2. Given this information we can now plug it into the following equation: F = m*a F = 4000kg * 2m/s^2 F = 8000kg * m/s^2 F = 8000 Newtons The opposing friction force exerts 5000 Newtons so we must add this to the Net Force in order to find the force that is applied. 8000N + 5000N = 13000 N The applied force is 1300 Newtons. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!