Assignment 10 qa

#$&*

course Phy 121

10/25 3

010. Note that there are 12 questions in this set.

Force and Acceleration.

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Question: `q001. If a block of mass 10 kg accelerates at 2 m/s^2, then what net force is acting on the block?

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Your solution:

Fnet = m * a

Fnet = 10 kg * 2 m/s^2

Fnet = 20 Newtons

confidence rating #$&*:

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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.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: `q002. How much force must be exerted by someone pulling on it to accelerate a 10 kg object at 2 m/s^2?

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Your solution:

Fnet = m * a

Fnet = 10 kg * 2 m/s^2

Fnet = 20 Newtons

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.

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Self-critique (if necessary):

I didn't think about all of the other factors, because values were not given other than mass and acceleration.

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Self-critique rating:OK

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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?

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Your solution:

Fnet = -10 Newtons + (10 kg * 2 m/s^2)

Fnet = -10 Newtons + 20 Newtons

Fnet = 10 Newtons

What we were looking for was just the Force of the 10 kg object moving 2 m/s^2, so therefore it would take 20 Newtons to make the 10 kg object move 2 m/s^2, with acceleration in the same direction as the motion.

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.

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Self-critique (if necessary):

Wow, I did not do it that way, but this way makes much more sense. I understand how this was done, but I am trying to really get a better grasp on it.

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Self-critique rating:3

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F_net = 10 kg * 2 m/s^2.

If F_applied is the force being applied, this means that since

F_applied + f_friction = F_net,

the equation would be

F_applied + (-10 N) = 10 kg * 2 m/s^2.

Solving this for F_applied, you would get the correct result, which is 30 Newtons.

*@

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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?

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Your solution:

Fnet = F + Ffrict

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.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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?

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Your solution:

Fnet = m * a

a = Fnet / m

a = 12 Newtons / 6 kg

a = 12 kg * m/s^2 / 6 kg

a = 2 m/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.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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?

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Your solution:

Fnet = F + Ffrict

Fnet = 50 Newtons + (-10 Newtons)

Fnet = 60 Newtons

Now we use Fnet = m * a

a = Fnet / m

a = 60 kg * m/s^2 / 20 kg

a = 3 m/s^2

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.

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Self-critique (if necessary):

I don't understand why you added it in question 4 as if they were both positive, but then subtracted it here acknowledging that one was negative and the other positive. I thought that since you got the Fnet in question 4 to be 30 Newtons, then here it would be 60 Newtons.

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Self-critique rating:3

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In Question 4 we were trying to find the applied force. We said that

20 Newtons = F + (-10 Newtons).

Here we are given the two forces that make up F_net, and we're trying to find the acceleration. We first find F_net by adding the two given forces, to get

F_net = 50 N + (-10 N) = 40 N.

*@

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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?

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Your solution:

Fnet = F + Ffrict

Fnet = (-50 Newtons) + (-10 Newtons)

Fnet = -60 Newtons

Now we find a

Fnet = m * a

a = Fnet / m

a = -60 kg * m/s^2 / 20 kg

a = -3 m/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.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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?

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Your solution:

I guess since we are given a value for velocity and need to find acceleration we also need to find `dt, which is ultimately what the question is asking.

Fnet = m * a

a = Fnet / m

a = -20 kg * m/s^2 / 40 kg

a = -0.5 m/s^2

Now that we know the a, and are given the velocity at this point, we can find `dt.

a = (vf - v0) / `dt

`dt = (vf - v0) / a

`dt = (0 m/s - 20 m/s) / -0.5 m/s^2 Since we are wanting it to come to rest that would make vf = 0

`dt = -20 m/s / -0.5 m/s^2

`dt = 40 s

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.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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?

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Your solution:

First we solve for a

a = (vf - v0) / `dt

a = (40 m/s - 10 m/s) / 5 s

a = 30 m/s / 5 s

a = 6 m/s^2

Now we need to find Fnet

Fnet = m * a

Fnet = 50 kg * 6 m/s^2

Fnet = 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.

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Self-critique (if necessary):OK

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Self-critique rating:OK

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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?

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Your solution:

We know the v0 = 8 m/s, `dt = 4 s, and vf = 0 m/s since it says that it comes to rest in the 4 seconds. From this we can find a.

a = (vf - v0) / `dt

a = (0 - 8 m/s) / 4 s

a = -8 m/s / 4s

a = -2 m/s^2

Now we can solve for m

Fnet = m * a

m = Fnet / a

m = 50 Newtons / -2 m/s^2

m = 50 kg * m/s^2 / -2 m/s^2

m = -25 kg.....but this would have to be positive 25 kg.

I know the mass can't be negative, but how would we know that the Fnet was a -50 and not a +50? Would we just assume this since the mass would be negative if the Fnet were positive, which cannot happen?

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.

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Self-critique (if necessary):

Oh I see now. It says that it must be in the same direction as the acceleration, therefore making it negative. I see, I had forgotten about that. I will not forget that again.

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Self-critique rating:3

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.

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Question: `q011. A system of mass 16 kg is subjected to opposing forces of 8 Newtons and 4 Newtons. What is its acceleration?

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Your solution:

I don't understand how I am supposed to know which of the forces is negative and the other positive, or if you disregard the negative, making it a total of 12 Newtons (in previous problems I have saw the instructor do both, and I don't know why he did it differently in each problem.)

Fnet = m * a

a = Fnet / m

a = 12 Newtons / 16 kg

a = 12 kg * m/s^2 / 16 kg

a = .75 m/s^2

confidence rating #$&*:

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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?

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Your solution:

We know that v0 = 0 m/s, vf = 10 m/s, and `dt = 5 s

a = (vf - v0) / `dt

a = (10 m/s - 0 m/s) / 5 s

a = 2 m/s^2

Therefore we know that the force of this object is positive since a is positive.

F = m * a

F = 4000 kg * 2 m/s^2

F = 8000 Newtons

Now that we know it has a positive force of 8000 Newtons along with a frictional force of 5000 Newtons, we can find how much force must be applied. (which I assume is disregardng the positives and negatives overall.)

Fnet = F + Ffrict

Fnet = 8000 Newtons + (-5000 Newtons)

Fnet = + 13000 Newtons

8000 Newtons in the direction of the object and 5000 Newtons opposite the direction of the moving object.

confidence rating #$&*:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#