# Work done by friction force on an object?

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1. Nov 24, 2014

### bluetriangle

I'm using a review book to study for physics but I don't understand how this answer was obtained. I got a different answer. When I googled the exact same question, I found a website that got the same answer as me... so is the book wrong? The question is as follows:

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

A 15 kg crate is moved along a horizontal floor by a warehouse worker who's pulling on it with a rope that makes a 30 degree angle with the horizontal. The tension in the rope is 200 N and the crate slides a distance of 10 m. If the coefficient of kinetic friction between the crate and floor is 0.4, how much work is done by the friction force?

The book uses 10 m/s² for gravity so for the sake of matching answers, I will too.

2. Relevant equations

W = Frcosθ

μ = Ff/Fn

3. The attempt at a solution

I drew a free body diagram. The only forces acting in the x-direction (direction of the displacement of the crate and therefore what is important to look at when solving for work) are Ftensionx (or Ftx) and Ff.

I know that Ftx = cos30 * 200 = 173.2 N
I know that Ff = μFn = μmg = 0.4*15*10 = 60 (I know Fn = Fg because the crate does not move in the y-direction)
I know that θ = 180 because Ff and the direction of displacement are antiparallel

W = Frcosθ = Ffrcosθ = 60*10*cos180 = -600 J

However, the book says -200 J
Am I doing this wrong?

2. Nov 24, 2014

### Staff: Mentor

There's a problem with your friction force. The force applied by the rope has two components, a horizontal one (which you got) and a vertical one. In which direction does this vertical component act on the crate? What does this do to the normal force which determines the friction force?

3. Nov 24, 2014

### bluetriangle

I don't understand exactly... The vertical force of the rope is pulling it upwards, combined with Ftx to pull it at an angle. But it doesn't do any work on the crate, correct? Because the crate is not moving vertically, only horizontally. That's why I only paid attention to the forces acting in the x-direction.

I thought Fn = Fg no matter what (if there is no movement in the y-direction) so I'm not connecting how Fty affects Ff.

4. Nov 24, 2014

### haruspex

No movement in the y direction tells you the forces balance in that direction, but Fn and Fg are only two of the three forces acting.

5. Nov 24, 2014

### bluetriangle

Oh! I see it now.

In the y-direction: Fn + Fty - Fg = 0
Fn = Fg - Fty = mg - mgsinθ
Fn = 50 (not 150)

So:
Ff = 50 * 0.4
Ff = 20 N

W = Frcosθ = 20*10*cos180 = -200 J

Yay thanks so much for your help! I really appreciate it :)