# Homework Help: Work & Power

1. May 27, 2005

### verd

Alright-- Sorry, I have a couple I'm stuck on...

...I'm still a little slow with this, so I make a lot of stupid mistakes-- If you see one, by all means, please let me know.
(Also, pardon the incredibly stupid nature of some of the problems... They're ridiculous)

1.) Your cat "Ms." (mass 7.00 kg) is trying to make it to the top of a frictionless ramp 2.00 m long and inclined upward at 30.0 degrees above the horizontal. Since the poor cat can't get any traction on the ramp, you push her up the entire length of the ramp by exerting a constant 100-N force parallel to the ramp.
If Ms. takes a running start so that she is moving at 2.40 m/s at the bottom of the ramp, what is her speed when she reaches the top of the incline? Use the work-energy theorem.

Okay-- So, moving down inclines I seem to have no problem with, but moving up inclines... eh... I seem to be having difficulty with. I've read somewhere that when you have a constant force, and you're moving up an incline, you just use the same old, W=Fs formula-- which makes sense. But this cat doesn't have a constant force-- Initially the cat does, when it gets a push, but I'm not quite understanding how I compare a constant force with one that isn't constant. (I do know and undersatnd the work-energy theorem.)
Needless to say, I've gotten tons of different answers on this one-- all of which are incorrect. Could anyone point me in the right direction on this one? Work and energy with varying forces?

2.) A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component $$F_x = -[20.0\;{\rm N} + (3.0\; {\rm N})x]$$.
How much work does the force you apply do on the cow during this displacement?

This one-- Work and energy with varying forces. I'm having a little difficulty with these, as I'm in Calculus right now... Eh, but I'm not here asknig for a tutorial on integrals, no worries. This:
$$W = \int F_xdx$$ will be equivalent to $$F_x(x_2-x_1)$$?
Is this the correct formula? ...How should I solve Fx?

3.)A block of ice of mass 4.50 kg is placed against a horizontal spring that has force constant k = 150 N/m and is compressed a distance 2.30×10−2 m. The spring is released and accelerates the block along a horizontal surface. You can ignore friction and the mass of the spring.
Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length.
What is the speed of the block after it leaves the spring?

This, too seems to be another of the same problem.
$$W = \int kxdx = 1/2kx_2^2 - 1/2kx_1^2$$?
...I initially tried solving this thinking that k2 was not = to 0 -- but apparently that's not correct. Any advice on how to go about this one?

Thank you ten times

2. May 27, 2005

### Staff: Mentor

Let's take them one at a time...
What makes you think the applied force is not constant? It says that it is. The work-energy theorem tells you that the work done by the hand pushing the cat must equal the change in mechanical energy of the cat. (Don't forget that both KE and gravitational PE will change.)

3. May 27, 2005

### HallsofIvy

At first I thought this was two separate situations. I assume that you cat first runs at the ramp and you push here with a 100 N force. Okay:
1) What was her kinetic energy at the bottom of the ramp?
2) How much work did you do applying 100 N for the 2.0 m length of the ramp?
3) What was her total energy at the top of the ramp?
4) What was her potential energy at the top of the ramp (relative to the bottom)?
5) What was her kinetic energy at the top of the ramp?
6) What was her speed at the top of the ramp?

Yes, $$W = \int F_xdx$$ is the work done. It is "equivalent" to W= F(x2- x1 in the sense that if F is a constant, the integral reduces to the product.
"How should I solve Fx?" If you mean "how should I solve for fx, you don't- you are given that in the problem. You need to do the integral- which should be simple.

4. May 27, 2005

### Staff: Mentor

cow problem

You are trying to find the work, which is given by $$W = \int F_xdx$$. You are given $F_x$, so plug it in and integrate from x = 0 to x = 6.9 m.

Note: This is not equivalent to $$F_x(x_2-x_1)$$; that would be true if $F_x$ were a constant, but it's not.

5. May 27, 2005

### Staff: Mentor

block of ice

You are on the right track. If you let the uncompressed position of the spring be at $x_2 = 0$, then you must integrate from x = -2.30×10−2 m to 0. Note that the force is F = -kx, not kx.

6. May 27, 2005

### verd

okay, wow. thanks-- i got the cow one down, I get it. THANK YOU!!

As for the weird one with the cat-- I haven't yet gotten to the chapter on potential energy / conservation of energy... However, the mechanical energy is equal to the kinetic + the potential-- so i'd have to find the potential for the bottom-- which, if the cat was moving, would be 0. But at the top? I can assume that the cat is still moving, right?
I realize that I'm supposed to use the work-energy theorem, but because the cat isn't moving at a constant force, by itself, but when it's being pushed, it is... I'm not sure how to compare the both. Am I still missing it? Am I wrong? Are they both constant? If so, how?

(Thank you Thank you, by the way)

7. May 27, 2005

### Staff: Mentor

the cat

Potential energy is just mgh and has nothing to do with speed. At the bottom of the ramp, you can call h = 0, so the gravitational PE is zero. At the top, you'd have to figure out the height it rose to.

But if you haven't covered that, no problem. Just find the net force on the cat parallel to the ramp. (Include the component of the cat's weight acting down.) If you do it that way you don't have to worry about PE, since you are including the force of gravity. Then the W-E theorem tells you that the work done by the net force equals the change in KE.

I don't know what you mean by "moving at a constant force". The force on the cat is constant! Of course the speed of the cat changes; that's what your trying to find out.

Think of it this way: At the bottom of the ramp the cat has an initial speed -- you don't care how it got that speed (it jumped!). The instant it landed on the ramp, you started pushing. So the force on the cat while it's sliding up the ramp is constant.

8. May 27, 2005

### verd

Okay, I think I'm getting it.

So, Friction was never included, and because we're looking for work, we don't need to include the normal force... So for the net force, I'm getting for the y-component -18.6, and for the x-component, 86.6... The magnitude of that ends up being 88.5749. ...So to get W, you have to take the dot product of the net force, and the straight-line displacement...

Fnet:
x- 86.6
y- -18.6

S
x- sqrt(3)
y- 1
Dot product: (86.6)(sqrt(3))+(-18.6)(1) = 131.396 J

And then comes the W-E theorem, plug in K1, find K2 with Wtot, and then you can solve for v...

Seem right?

[By the way... You really really helped me get a lot of this. THANK YOU TIMES TWENTY.]

9. May 27, 2005

### Staff: Mentor

I'm not sure what you're doing here. Since the cat only moves parallel to the ramp, all we need consider are forces parallel to the ramp. There are two: The applied force of 100 N up the ramp and the component of the weight acting down the ramp. Combine them to get the net force parallel to the ramp. Then use the work-energy theorem to find the change in KE.

10. May 29, 2005

### verd

Yep-- that's what I did... By net force, I meant the combination of gravity/weight, and the 100N force pushing horizontally up the ramp. Got the answer right, thank you for all of your help.

I have another question that I can't seem to understand the wording to-- I can answer the question, assuming I understand exactly what's going on. Here it is:

A sack of flour of mass 3.50 kg is lifted vertically at a constant speed of 3.20 m/s through a height of 10.0 m.
They then ask you to calculate the force required, which I got as 34.3N, and the amount of work done onthe sack by the lifting force, which I got as 343J.
What becomes of this work?

By that, do they mean after it has reached the 10m height? Has it stopped? Or is it still being lifted? They give you the following choices:

# This work becomes kinetic energy.
# This work becomes potential energy.
# This work becomes both kinetic and potential energy.
# This work becomes heat.

...Like I said, I can definatley answer the question-- I was just wondering what that wording meant to someone else...

11. May 29, 2005

### Staff: Mentor

A certain amount of work had to be done to lift the sack. That means energy was transferred to the sack. What happened to it?

So... yes they mean after it reaches 10m. No information was given about what happens after the sack reaches 10m - whether it is then stopped (by some other force) or keeps moving because the force is still being applied. But that doesn't matter. All that matters is: What happened to the work that was done in lifting the sack up to that height? I think you should be able to answer the question now.