# Work done by a constant force question

1. Oct 9, 2009

### farleyknight

1. The problem statement, all variables and given/known data

A 9.9 kg body is at rest on a frictionless horizontal air track when a constant horizontal force $$\vec{F}$$ 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 Fig. 7-26. The force $$\vec{F}$$ 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 $$\vec{F}$$ between t = 0.71 and t = 1.1 s?

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c07/fig07_26.gif

2. Relevant equations

3. The attempt at a solution

Firstly, unless my numbers are wrong, the acceleration for object this is not constant.

From t = 0 to t = 0.5, $$\delta s = 0.04$$. So then $$v = 0.08$$

From t = 0.5 to t = 1.0, $$\delta s = 0.16$$. So then $$v = 0.24$$

From t = 1.0 to t = 1.5, $$\delta s = 0.24$$. So then $$v = 0.48$$

But notice that for the first two intervals, $$\delta v = 16$$ while for the second, it is $$\delta = 0.24$$. So the acceleration is not constant on the interval t = 0 to t = 1.5.

However, I did manage to boil down the answer to the solution to this equation:

$$d = (x_f - x_i) = \frac{1}{2}a*t_f^2 - \frac{1}{2}a*t_i^2 = \frac{1}{2}a*(t_f^2 - t_i^2)$$

Solving for work:

$$W = Fd \cos(\phi) = m*a*d = \frac{1}{2}m*a^2*(t_f^2 - t_i^2)$$

Subbing the values:

$$W = (0.5)*(9.9)*(1.1^2 - 0.71^2)*a = 3.4947*a$$

Where a is whatever value that works, given the data. Note that I've tried a = 0.48, the acceleration that I found from t = 0 to t = 1.5, a = 0.32 which is the discrete average acceleration from t = 0 to t = 2.0, a = 0.426, which is the arithmetic average of the three values for a I have. Suffice to say none of them work.

Kinda feel dumb for being hung up on such a simple problem, but whatever..

2. Oct 9, 2009

### Vykan12

If the force applied and external forces are all constant, what gives you the impression that acceleration isn't?

3. Oct 9, 2009

### farleyknight

The data! I should post up the chart I made to show you what I mean, but I don't know if the forums accept tables.

4. Oct 9, 2009

### Vykan12

I plotted the points on my calculator and got the linear expression a = 0.4v - 0.1333333...

This seems like a strange problem to give as homework. Also, how did you come up with that expression for acceleration?

5. Oct 9, 2009

### farleyknight

There are 5 data points, which gives 4 position deltas:

$$\delta s_1 = 0.04 - 0 = 0.04$$
$$\delta s_2 = 0.20 - 0.04 = 0.16$$
$$\delta s_3 = 0.44 - 0.20 = 0.24$$
$$\delta s_4 = 0.80 - 0.44 = 0.36$$

Which in tern gives you 3 velocity deltas: (using *2 instead of / 0.5)

$$\delta v_1 = 2*(0.16 - 0.04) = 0.24$$
$$\delta v_2 = 2*(0.24 - 0.16) = 0.16$$
$$\delta v_3 = 2*(0.36 - 0.24) = 0.24$$

And from there you can just * 2 again for the average velocity on that interval.

$$a_1 = 2*0.24 = 0.48$$
$$a_2 = 2*0.16 = 0.32$$
$$a_3 = 2*0.24 = 0.48$$

From there, I decided to try out 0.48 since it included t = 1.1 and t = 0.71. Then I tried to find the acceleration between the first and last velocities

$$a_{avg} = \frac{v - v_0}{t - t_0} = \frac{0.72 - 0.8}{2} = 0.32$$

When that didn't work, I tried averaging the 3 values together to get:

$$a = \frac{0.48 + 0.32 + 0.48}{3} = 0.426$$

These are the three that were provided by the text book, although it's possible I overlooked one, but I doubt it.

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