- #1

thehammer

- 15

- 0

## Homework Statement

A 3kg body is at rest on a frictionless horizontal air track when a constant horizontal force F acting in the positive direction of an x-axis along the track is applied to the body. The force F is applied to the body at t = 0 when the body is at the origin. A stroboscope records the position of the body at 0.5 s intervals. The readings are as follows.

Interval (I) 1: 0.04 m

I2: 0.2 m

I3: 0.44 m

I4: 0.8 m

How much work is done on the body by the applied force F between t = 0 and t = 2s?

## Homework Equations

K = 1/2mv^2

W = [tex]\Delta[/tex]K

s = ut + 1/2at^2

v^2 = u^2 + 2as

## The Attempt at a Solution

I really have no idea why I'm not getting the right answer. I can't see an error in what I've done.

1) Find a

u = 0

a = 2s/t^2

a = 2 x 0.04 / 0.5^2 = 0.32 ms^-2

2) Find v^2

v^2 = ((0 ms^-1)^2) + 2(0.32 ms^-1 x 0.8m)

3) Find the change in kinetic energy and hence the work done

[tex]\Delta[/tex]K = 1/2mv^2 = 1/2(3kg)(0.512m^2 s^-2) = 0.768 J

---

The answer is meant to be 0.96J. I think I might have done this question months ago but I cannot see where I have gone wrong in the least. Please point out any glaring mistakes :P.