How Is Work Calculated on a Mass on a Frictionless Track?

[lemonpie]

Homework Statement

A 3.0 kg body is at rest on a frictionless horizontal air track when a constant horizontal force 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 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 between t = 0 and t = 2.0 s?
http://www.lowellphysics.org/beta/Textbook%20Resources/Chapter%207.%20Kinetic%20Energy%20and%20Work/Problems/c07x7_11.xform_files/nw0308-n.gif

Homework Equations

um... v = x/t, a = v/t, F = ma, W = Fd?

The Attempt at a Solution

delta v = x/t = 0.8/2 = 0.4 m/s
delta a = v/t = 0.4/2 = 0.2 m/s^2
F = ma = 3*0.2 = 0.6 N
W = Fd = 0.6*0.8 = 0.48 J

This is not the correct answer. The correct answer is 0.96 J, twice my answer. What am I doing wrong? Please help me. This is not an assigned problem for points; it's just to help me understand.

 

[turin]

You know that the force is constant. What does that tell you about acceleration? Once you figure that out, then determine which of the other kinematical variables that you know, and which ones you do not know, from the list: t_i, t_f, x_i, x_f, v_i, v_f. From this, and what you determine about the acceleration, you should be able to pick the appropriate kinematical equation.

 

[lemonpie]

You know that the force is constant. What does that tell you about acceleration?

I think it tells me... that the acceleration is also constant. Should it tell me something else?

As for the variables:
ti = 0 s
tf = 2 s
xi = 0 m
xf = 0.8 m
vi = ?
vf = ?

I suppose if the acceleration is constant... then I can use those formulas in the beginning of the book... but I don't feel like that helps me, because all of those formulas seem to require vi or a, neither of which I have. Oh wait, vi = 0. So in that case...

vi = 0 m/s

xf - xi = 1/2(vi + v)t
0.8 = 1/2(2)v
v = 0.8 m/s

Then

v = vi + at
0.8 = 2a
a = 0.4 m/s^2

Then

F = ma
F = 3(0.4) = 12 N

Then

W = Fd
W = 12(0.8) = 0.96 J

Awesome! Thanks! I wonder why I couldn't just use the v = x/t... oh well. Thanks again!

 

[turin]

I wonder why I couldn't just use the v = x/t...

Because v is not constant. That equation is only valid for constant (or average) speed.

BTW, there may have been a more direct approach (i.e. single kinematical equation), but I don't know what is in your book. Anyway, it just takes a while to get the hang of when to use which formula.

 

[lemonpie]

Because v is not constant. That equation is only valid for constant (or average) speed.

Oh, right.

I'm sure there was a more efficient way to solve this problem, but I guess the thing for now was just for me to be able to solve it! TQ