Simple Work Physics Problem?

Question

A 2.5 kg body is at rest on a frictionless horizontal air track when a horizontal force F acting in the positive direction of an x axis along the track is applied to the body. A stroboscopic graph of the position of the body as it slides to the right is shown in Figure 7-26. The force F is applied to the body at t = 0, and the graph records the position of the body at 0.50 s intervals. How much work is done on the body by the applied force F between t = 0 and t = 1.0 s? (HINT: Find the average acceleration over t = 0 to t = 1.0 s.)

Here's my best attempt at drawing the graph.

http://i89.photobucket.com/albums/k239/blueray4/physicsprob.jpg

I thought this was so simple, but my answer is being rejected for some reason. Here's my work.

a= (.2m/s-0m/s)/1s=.2m/s^2

F=ma
F=2.5kg *.2m/s^2
F=.50N

W=F*d
W=.50N * .2m
W=.10J

what the heck? that should be right? or am I doing something stupid?

Answer 1

Given:

s=0.2m based on the graph.
t=1s
u=0

Solve for acceleration, a, using the formula:

s=ut+1/2at^2

where s= the distance traveled
u=the initial velocity
a=acceleration
t=time

Substitute given values:

0.2=0*1+1/2a*1^2
0.2=1/2a
a=0.4m/s^2

Now solve for F using the formula:

F=Ma

where F is the applied force, M the mass, and a the acceleration.

Given:

M=2.5kg
a=0.4m/s^2

Substitute given values:

F=2.5*0.4
=1N

Solve for the work using the formula:

Work=F*s

where F is the applied force and s is the distance traveled.
Given:
F=1N
s=0.2m

Substitute given values:

Work =1*0.2
=0.2 joule

Note: the first equation in your solution is incorrect. 0.2m as given in the graph is the distance traveled. It is not equal to acceleration. That's why we have to solve for the acceleration as our first step in solving the (simple) problem.