Work done by a constant force question

AI Thread Summary
A 9.9 kg body on a frictionless air track is subjected to a constant horizontal force, but the acceleration appears to vary based on the position data collected at specific time intervals. Calculations show discrepancies in acceleration values, leading to confusion about the work done by the force between t = 0.71 and t = 1.1 seconds. Despite attempts to derive a consistent acceleration value using different methods, none yield satisfactory results. The discussion highlights the complexity of the problem, questioning the assumption of constant acceleration when data suggests otherwise. The challenge lies in reconciling the observed data with theoretical expectations of motion under constant force.
farleyknight
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Homework Statement



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

Homework Equations





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..
 
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If the force applied and external forces are all constant, what gives you the impression that acceleration isn't?
 
Vykan12 said:
If the force applied and external forces are all constant, what gives you the impression that acceleration isn't?

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.
 
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?
 
Vykan12 said:
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?

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 textbook, although it's possible I overlooked one, but I doubt it.
 
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