# Work done by friction pushing blocks

• preluderacer
In summary, a constant external force of 170 N is applied to a 20-kg box 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 changes the speed from 0.4 m/s to 3.0 m/s. Using the work-energy theorem, the work done by friction is calculated to be 1089.39 J, with a negative value indicating that the friction force acts opposite to the direction of displacement. The given time interval of 9.0 s is found to be incorrect, as using it in the calculations yields different results.
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.

Use the work-energy theorem in the horizontal direction.

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

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

There seems to be an error in the problem statemnt.

What do you mean?

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.

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

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

Is W_f positive or negative?

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

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.

## What is work done by friction pushing blocks?

Work done by friction pushing blocks refers to the energy expended when an object is moved by the force of friction against another surface. This energy is typically dissipated as heat.

## How is work done by friction calculated?

The work done by friction can be calculated by multiplying the force of friction by the distance over which the object is moved. This is represented by the equation W = Fd, where W is work done, F is the force of friction, and d is the distance moved.

## What factors affect the amount of work done by friction pushing blocks?

The amount of work done by friction pushing blocks is affected by several factors including the force of friction, the type of surface the object is moving on, and the weight and speed of the object being pushed.

## Can work be done by friction pushing blocks in a vacuum?

No, work cannot be done by friction pushing blocks in a vacuum since there is no air or other medium for friction to act upon.

## How does the direction of force affect the work done by friction pushing blocks?

The direction of force does not affect the work done by friction pushing blocks. As long as there is movement and friction between two surfaces, work will be done regardless of the direction of the force.

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