#$&* course PHY 121 6/24 3There are a bunch of problems under these that you usually ask us to do and not turn in. That line is not above them, so right now I'm doing them. I'll turn them in if you need them. Question: `q001. If a block of mass 10 kg accelerates at 2 m/s^2, then what net force
.............................................
Given Solution: The net force on the block is the product F = m * a of its 10 kg mass and its 2 m/s^2 acceleration. The net force is therefore F = 10 kg * 2 m/s^2 = 20 kg * m / s^2. The unit of force, which is the product of a quantity in kg and another quantity in m/s^2, is just the algebraic product kg * m/s^2 of these two units. This unit, the kg * m / s^2, is called a Newton. So the net force is 20 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK I wasn't sure if the final unit would be Newtons ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. How much force must be exerted by someone pulling on it to accelerate a 10 kg object at 2 m/s^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Force = mass times acceleration F = 10 kg * 2 m/s^2 F= 20 Newtons confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 did not consider for more than a moment all of the ""what ifs"". ------------------------------------------------ Self-critique rating: 2 ********************************************* 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: Without taking the friction into consideration, it would be 10 kg * 2 m/s^2 = 20 Newtons. I am assuming that, since the friction is exerting a force of 10 Newtons in the opposite direction, the person must overcome that by exerting 10 more Newtons of force for a total of 30 Newtons. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Since the 10 Newton frictional force is in the direction opposite to motion, and since the acceleration is in the same direction as the motion, the frictional force is opposed to the accelerating force. If the direction of motion is taken as positive, then the frictional force will be in the negative direction and can be denoted fFrict = - 10 Newtons. To achieve the given acceleration the net force on the object must be net force = 10 kg * (+2 m/s^2) = +20 kg * m/s^2 = +20 Newtons. In order to achieve the +20 Newton net force when there is already a frictional force of -10 Newtons, it should be clear that a force of +30 Newtons is required. This result can be interpreted as follows: The person must exert 10 Newtons of force to overcome friction and another 20 Newtons to achieve the required net force. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK I just need to learn the scientific formul approach. ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: `q004. How can we write an equation to solve this problem? Hint: What equation would relate the net force Fnet, the force F exerted by the person and the force fFrict of friction? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet = F + fFrict The total force exerted is the force exerted by the person and the force of Friction. I'm assuming in the previous problem, it would be solved as: 10 kg * 2 m/s^2 = 20 Newtons is the Fnet. F was what we were trying to find. And the force of friction would be -10 Newtons because it was working against the person's efforts. So you would have 20 Newtons = F + (-10 Newtons). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 Thought I'd only gotten maybe a piece of that correct. ------------------------------------------------ Self-critique rating: 3 ********************************************* 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: a = Fnet / m a = 12 Newtons / 6 kg 12 Newtons is the same as 12 kg/ m/s^2 so a = 2 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The velocity of the cart will change at a rate a which is related to the net force and the mass by Fnet = m * a. Thus a = Fnet / m = 12 Newtons / (6 kg) = 12 kg * m/s^2 / (6 kg) = 2 m/s^2. We note that the force unit Newtons is broken down to its fundamental units of kg * m / s^2 in order to perform the unit calculation. Dividing kg * m / s^2 by kg we have (kg / kg) * m/s^2 = m/s^2. It is important to always do the unit calculations. This habit avoids a large number of errors and also can be used to reinforce our understanding of the relationships in a problem or situation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: `q006. If a force of 50 Newtons is exerted in the direction of the object's motion by a person, on a 20 kg object, and if friction exerts a force of 10 Newtons opposed to the direction of motion, then what will be the acceleration of the object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First you have to find the net force Fnet = F + fFrict Fnet = 50 Newtons + (-10 Newtons) Fnet = 40 Newtons Acceleration is found by a = Fnet / m a = 40 Newtons / 20 kg 40 Newtons is the same as 40 kg / m/s^2 a = 2 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: 3 ********************************************* 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 F is - 50 Newtons because the person is exerting 50 Newtons of force in the opposite direction of the object's motion fFrict is - 10 because friction is exerting 10 Newtons of force in the opposite direction of the object's motion This make the Fnet = -50 Newtons + (-10 Newtons) or -60 Newtons Acceleration is Fnet divided by mass a = -60 Newtons/20 kg -60 Newtons is -60 kg / m/s^2 a = -3 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: If we take the direction of motion to be positive, then since both the 50 Newton force and the 10 Newton force are opposed to the direction of motion the net force must be net force = -50 Newtons - 10 Newtons = -60 Newtons. The acceleration of the object will therefore be a = Fnet / m = -60 Newtons / (10 kg) = -60 kg * m/s^2 / (20 kg) = -3 m/s^2. The fact that the acceleration is opposed to the direction of motion indicates that the object will be slowing down. The force exerted by the person, being in the direction opposite to that of the motion, is seen to be a retarding force, as is friction. So in this case the person is aided by friction in her apparent goal of stopping or at least slowing the object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q008. If a 40 kg object is moving at 20 m/s, then how long will a take a net force of 20 Newtons directed opposite to the motion of the object to bring the object to rest? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Fnet = -20 Newtons Mass = 40 kg. a = Fnet / m a = -20 Newtons / 40 kg a = -.5 m/s^2 a = `dv / `dt `dt * a = `dv `dt = `dv / a `dt = (vf - v0) / a `dt = (0 m/s - 20 m/s) / -.5 m/s^2 `dt = 40 s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: 2 ********************************************* 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: First I found the acceleration:40 m/s - 10 m/s / 5 s = 30 m/s / 5 s = 6 m/s^2 Then I used Fnet = a * m Fnet = 6 m/s^2 * 50 kg Fnet = 300 Newtons confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The net force would be Fnet = m * a. The acceleration of the object would be the rate which its velocity changes. From 10 m/s to 40 m/s the change in velocity is +30 m/s; to accomplish this in 5 seconds requires average acceleration 30 m/s / (5 s) = 6 m/s^2. Thus the net force required is Fnet = m * a = 50 kg * 6 m/s^2 = 300 kg m/s^2 = 300 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK As long as I have the examples and formulas in front of my, I'm able to plug the numbers in and get the answers. ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: `q010. If a constant net force of 50 Newtons brings an object to rest in four seconds from an initial velocity of 8 meters/second, then what is the mass of the object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I started with a = `dv/`dt and `dv - 0 m/s - 8 m/s, so -8 m/s / 4 s = -2 m/s^2 for acceleration Then I went to Fnet = a * m. This means m=Fnet/a Because the force was bringing the object to rest, I went with it being a negative -50 Newtons for the force. So -50 Newtons / -2m/s^2 = 25 kg confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: 2 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: I am interpreting that the mass has two forces working against it, rather than against each other. If this is the case: a = Fnet/m; a = (-8 Newtons - 4 Newtons) / 16 kg. This would give you -12 Newtons/16 kg, or .75 m/s^2 If it means that the forces are opposing each other, then you would either have positive 4 or negative 4 working on the mass. This would give you either .25 m/s^2 or -.25 m/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q012. A 4000 kg mass is to accelerated uniformly from rest to 10 meters/second during a 5-second interval. It experiences an opposing frictional force of 5000 Newtons. How much force must be applied? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Acceleration is (10 m/s - 0 m/s) /5 m/s = 2 m/s^2 Fnet = a * m 2 m/s^2 * 4000 kg = 8000 Newtons Fnet = F + fFrict 8000 Newtons = F + (-5000 Newtons) F = 13000 Newtons of force need to be applied confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Self-critique (if necessary): ------------------------------------------------ Self-critique rating: