# Work done by a force on a body

1. Feb 11, 2006

### Amith2006

Sir,
Under the action of a force, a 2kg body moves such that its position X as a function of time t is given by X = t^3/3 where X is in metres and t in seconds. What is the work done by the force in the first 2 seconds? Here the symbol ^ denotes power. Please help me in solving this problem.

2. Feb 11, 2006

### siddharth

What have you attempted with this problem? You need to show your work, because it is easier to help that way.

3. Feb 12, 2006

### Meir Achuz

1. Find F=m(d^2x/dt^2).
2. Write F in terms of x.
3. Integrate Fdx.

4. Feb 12, 2006

### Amith2006

Sir,
I solved it the following way :-
Now X = t^3/3
Differentiating the above equation w.r.t time we get,
dX/dt = v = t^2 ( v denotes velocity)
Differentiating the above equation w.r.t time again we get,
d^2X/dt^2 = a = 2t (a denotes acceleration)
a(t = 0) = 0
a(t = 2) = 4 m/s^2
Average a = (0+4)/2 = 2 m/s^2
X( At t = 0) = 0
X(at t = 2) = 8/3 = 2.67 m
Displacement in the first 2 seconds = 2.67 m
Work done = Force x Displacement
= mass x acceleration x Displacement
= 2 x 2 x 2.67
= 10.68 Joules

But the answer given in my book is 16 Joules. I solved a similar kind of problem but in that problem X = 2t^2. So i got the acceleration as a constant on differentiation.Could you please explain it to me if it is wrong?

5. Feb 12, 2006

### Amith2006

Sir,
I did as you said. I got the answer as 10.98 Joules. Is it right?

6. Feb 12, 2006

### Cyrus

Your work seems fine until you averaged it. I would have said,

$$w= \int F * dx = \int (ma)dx = \int^2_0 (2)(2t)(t^2 dt) = 16 Joules$$

Because, $$dx = t^2 dt$$

7. Feb 12, 2006

### finchie_88

This is probably not the best way, but this is how I did it.
$$x = \frac{1}{3}t^3$$
$$W = \int F.dx$$
Where, $$F = ma = m\frac{d^2x}{dt^2}$$, so, $$\frac{d^2x}{dt^2} = 2t$$
[tex] W = \int 2mt^3.dt [tex]
Then solve that.
edit: dam, beaten to it.

Last edited by a moderator: Feb 12, 2006
8. Feb 12, 2006

### Cyrus

Didn't I just post that? No, you were about 33 mins behind me.

Last edited: Feb 12, 2006
9. Feb 12, 2006

### finchie_88

yer, just u beat me to it, looking at the times,. you beat me by quite a lot.

10. Feb 14, 2006

### Amith2006

Thank you very much Sir.

11. Feb 14, 2006

### vaishakh

The easiest method to do this is work done = change in Kinetic energy