- #1
thehammer
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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
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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.