Work done by friction pushing blocks

[preluderacer]

Homework Statement

In the figure, a constant external force P = 170 N is applied to a 20-kg box, which is on a rough horizontal surface. The force pushes the box a distance of 8.0 m, in a time interval of 9.0 s, and the speed changes from ν1 = 0.4 m/s to ν2 = 3.0 m/s. The work done by friction is closest to:
(P is pushing down at a 30 degree angle in the picture)

The Attempt at a Solution

I know I need to find the horizontal component of the push, but I don't know what to do next.

 

[PhanthomJay]

Use the work-energy theorem in the horizontal direction.

 

[preluderacer]

Im not quite sure were the 9 seconds fits into this?

 

[PhanthomJay]

Im not quite sure were the 9 seconds fits into this?

There seems to be an error in the problem statemnt.

 

[preluderacer]

What do you mean?

 

[PhanthomJay]

If you work out the problem using the data that it travels 8m with the initial and final velocities as noted, you get the correct answer, and the acceleration is constant. If you work out the problem using the fact that the time is 9 seconds, you get a completely different acceleration, so the time given is incorrect. Are you familiar with work energy methods? If not, fine, try the problem using the other data and the kinematic equations and Newton's laws, and see that you get different results based on the 2 sets of data.

 

[preluderacer]

Ok so here's what I did I calculated the force in the applied to the box to be 170cos30=147.22N. After that I multiplied that by 8 meters and got that the work done in that direction was 1177.79 J. Next I used the work energy theorem from ν1 = 0.4 m/s to ν2 = 3.0 m/s and got 88.4 J. I then subtracted 1177.79 J - 88.4 J and got 1089.39 J. Now is that final answer the work done by the friction? If so, is that work negative or positive

 

[PhanthomJay]

W_P + W_f = delta K = 88 J
1178 J + W_f = 88 J

Is W_f positive or negative?

 

[preluderacer]

It is negative because its in the opposite direction of the pushing force?

 

[PhanthomJay]

The math shows it is negative, which implies that work is negative because the friction force acts opposite fo the direction of the diplacement . Work is the dot product of foce and displacement , W_f = f.d =f(d)(cos theta), where theta is the angle between the force and displacement vectors. The friction force acts to the left, and the displacement is to the right, so in this case, work = fdcos 180, = f(d)(-1) = a negative number.