Another Work Problem

1. Oct 22, 2008

Bones

1. The problem statement, all variables and given/known data

http://www.webassign.net/userimages/jlabelle@dartmouth/graph.gif

A block with mass m = 10 kg slides along a horizontal floor. The kinetic friction force opposing the motion of the block is constant at a value 14 N. At clock reading t=0.0 s and position x=0.0 m, the block has an instantaneous velocity of v = 4 m/s to the right. A force P is applied to the block starting at the instant it is in the position x=0. The direction of P remains fixed and positive (to the right), but its magnitude varies according to the graph above. Use work and energy arguments to answer the questions below; do not make use of Newton's second law.
a) Calculate the work done by the force P in the displacement from x=0 to x=4 m.

b) Calculate the work done by the frictional force during the same displacement

c) Calculate the work done by the net force acting on the block during the displacement.

d) Calculate the change in kinetic energy of the block.

e) Calculate the final kinetic energy of the block; that is, the kinetic energy it possesses on reaching the position x=4 m.

f) Calculate the velocity of the block at the position x=4 m.

g) Calculate how far the block will slide beyond position x=4 m if the force P abruptly drops to zero at this position and remains zero from there on.

h) Calculate any potential energy changes of the earth-block system that take place over the course of the motion from x=0 to x=4 m.

2. Relevant equations

3. The attempt at a solution
a) I found the area under the curve = 100 J
b) 14N*4m*cos180 = -56 J
c or d, I am not sure which) 100J+(-56J) = 44 J
I am not sure about the rest. Can someone help me out??

Last edited: Oct 22, 2008
2. Oct 22, 2008

LowlyPion

d) what happened to the additional work? What did it go into?

e) what's the total?

f) You know m so what's v?

g) how long until it stops then with friction slowing it by 14 J per meter?

h) is it horizontal?

3. Oct 22, 2008

Bones

Is the answer for c and d the same??

4. Oct 22, 2008

LowlyPion

Where else would it go?

5. Oct 22, 2008

Bones

I am not sure I know what part e is asking for...would initial kinetic energy be 1/2(10kg)(4m/s)^2=80 and then final would be 80-44=36??

6. Oct 22, 2008

Bones

For part f, 44J=1/2(10kg)v^2-1/2(10kg)(4m/s)^2 = 4.98m/s ?

7. Oct 22, 2008

Bones

8. Oct 22, 2008

Bones

g) 1/2(10kg)(4.98m/s)^2 = (14N)(x) x = 8.86m ?

Last edited: Oct 22, 2008
9. Oct 22, 2008

Bones

Sorry to be such a bother, but I applied for a tutor but they did not have one that could work around my school schedule :P

10. Oct 22, 2008

LowlyPion

P was adding KE to the object, not subtracting it.

11. Oct 22, 2008

Perillux

You found the initial KE correctly, but not the final.

Wnet = change in kinetic energy = KEfinal - KEinitial
Wnet = 44J as you found out, so:
44J = KEfinal - 80J
KEfinal = 44J + 80J = 124J

This looks good.

for part h, I'm not sure, but I would guess that the answer is zero.

For part g, there is still a frictional force of 14N slowing it down. But remember, you said you are not allowed to use newtons second law (F = ma) so you have to use the Wnet = KE2 - KE1 to solve for it.

12. Oct 22, 2008

LowlyPion

No. The total KE at that point is equal to mV2/2

Edit: OK. You got the right answer, but I was looking for 124 J = mV2/2

Last edited: Oct 22, 2008
13. Oct 22, 2008

Bones

So it would be 80+44?

14. Oct 22, 2008

Bones

So 124J=1/2(10kg)(v^2) v=4.98m/s

15. Oct 22, 2008

LowlyPion

This is the right idea, but the wrong total KE at x=4.

Just redo it.

Edit: OK so you did have the right total KE. I didn't recognize it.

16. Oct 22, 2008

Bones

KE at x=4 is 124J
124J=(14N)(x)
x=8.86m

17. Oct 22, 2008

Bones

That was a lot of great help. Thank you soo much!!