# Work done by a force slowing an object

1. Sep 27, 2013

### Mikry

This is one area of basic physics that I feel I'm never sure what is wanted in the question: what is the work done by a force decelerating an object?

I mean work is F*d, but in this case the distance moved by the object is in the opposite direction of the force. I mean I would think the work done by the force would then be the change in kinetic energy, which makes sense to me, but the problem is that in a similar question that came up in an exam I was marked wrong for that, here's what the question was more or less:

A ball of mass m is thrown in the air(negligible air resistance) with velocity v from height h1 off the ground. It reaches height h2 then falls back down to h1. What is the work done by gravity on the ball?

What I figured is that the work done would be the gain in GPE (or loss in KE) to get it to its maximum height, plus the gain in KE (or loss of GPE) for it to fall back to its original height. Thus I would have given my answer as 2mgh.

However, apparently the answer was zero. I mean I guess we could say it's because work is the conversion of energy from one form to another, and at the end of the sequence of events the ball has the same KE and GPE as it had at the start. But looking at it from that perspective in my mind has a very vector-like nature to it (eg the difference between distance and displacement), and work is a scalar property.

Additionally(though not entirely relevant in my original question), the accompanying diagram of the events in the question showed that the ball wasn't thrown straight up, but at an angle, thus there was even a horizontal distance moved. Personally I think the question was flawed, but you try telling my teacher that!

2. Sep 27, 2013

### tiny-tim

Hi Mikry!
definition: W = F.d

on the way up, the force is down and the displacement is up, so the work done is negative

on the way down, the force is down and the displacement is up, so the work done is positive

total zero!
since horizontal displacement is perpendicular to the force, it makes no contribution to the dot-product

3. Sep 27, 2013

### Mikry

Oh wow, now I feel a little silly!

Though I didn't know that work was the dot-product of displacement and force, I was always taught that it was the force multiplied by the distance moved. Thank goodness I learnt the correct definition before I ran into more trouble later on!

Just to confirm I understand how this all fits together, say I have an object in a 2D plane, that is displaced from point (0, 0) to point (x1, y1) by a force (a, b) (that's supposed to be in matrix notation but I don't know how to do that on this site...but component a is in direction x and component b is in direction y.), the work done will be (x1*a)+(y1*b). Have I got it right?

4. Sep 27, 2013

### tiny-tim

Hi Mikry!
Yup!

(and if the force varies, or if the path is not straight, then it's ∫ F·ds, the integral of force "dot" an infinitesimal distance ds along the curve)

5. Sep 27, 2013

### Mikry

That sounds quite interesting(and I little out of the scope of what I'm expected to know, but that's not important :tongue:), though I'm not sure how one would calculate the work done in that situation...I mean I understand what you're doing, but if I were given a set of relevant data I wouldn't really know how to plug it into that formula to get the value for the work done...

6. Sep 27, 2013

### tiny-tim

eg, if it's gravity, then F is a constant, -mgy,

and if the path is a curve (eg a parabola! ),

then ∫ F.ds = -mg ∫ y.ds,

which if you think about it is simply -mg(yf - yi)

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