Find work done by force given distances and times

1. Dec 8, 2015

dsb1713

1. The problem statement, all variables and given/known data
A 1.0 kg body is at rest on a friction-less 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. A stroboscopic graph of the position of the body as it slides to the right is shown in (the) figure. The force F is applied to the body at t = 0, and the graph records the position of the body at 0.5 second intervals. How much work is done on the body by the applied force F between t = 0 and t= 2.0 seconds?

At 0.5 seconds, the distance is 0.04 meters, at 1 second it is 0.2 meters, at 1.5 seconds it is 0.44 meter, at 2 seconds it is 0.8 meters.

2. Relevant equations
Work is equal to the change of kinetic energy. W=Kf-Ki.
Kinetic energy is equal to one half the mass times instantaneous velocity squared. K=1/2 mv^2
Displacement is equal to one half times initial velocity plus final velocity, multiplied by the time. Xf-Xi=(Vi+Vf)/2 *t
3. The attempt at a solution
I am doing this exercise in order to practice for our final test. I've found a solution, but my professor refuses to provide us with an answer key to the practice problems he gave us, so I cannot check whether my work is correct. I've found similar problems online but none with my exact numbers.

Solution attempt:
Because the object starts from rest, I assume the initial kinetic energy will be zero. Thus, because work is the change in kinetic energy, which is Kf minus Ki, it should be equal solely to the final kinetic energy. Therefore, we must find Kf.
Kf=1/2mv^2
We lack the final velocity Vf. However, given that we know the displacement and the time, we can find it by isolating the Vf variable from the following equation: Xf-Xi=(Vi+Vf)/2 *t
Vf will be equal to 0.8m/s
Plugging it into the previous Kinetic energy function, we get a total work of 0.32J.
Is my response correct? Is there a better approach?

2. Dec 8, 2015

Staff: Mentor

Hi dsb1713, Welcome to Physics Forums.

I've changed your thread title to: "Find work done by force given distances and times". This is to comply with forum rules that require thread titles to be descriptive of the thread content and the nature of the problem.

What definition do you know for work (other than the change in KE one, which is the work-energy theorem).

3. Dec 8, 2015

dsb1713

Thank you for adjusting the title.
I have Work being defined as the product of a Force vector and a Displacement vector. Also, work is equal to the negative change in potential energy.
In this case, I believe we could use the first of these two, using 0.8 meters as our displacement. For the force, using the standard F=m.a equation, I would need to find the acceleration using 2D kinematics formulas.
Am I on the right track?

Edit: Through 2D kinematics I found the acceleration to be 0.4m/s^2. By multiplying it with the mass, I found the force to be 0.4N. Now, by multiplying the force by the distance, I found the Work to be 0.32 Joules, consistent with my previous answer. Thank you so very much!

4. Dec 8, 2015

Staff: Mentor

Yes, looks like you're on the right track. In fact your initial approach is also good. There are several ways to get to the desired goal.

If this were a real lab where you took measurements you might want to plot a graph of the collected data in a way that you could draw a best fit line to extract the acceleration, then determine the force using f = ma.

5. Dec 8, 2015

Staff: Mentor

I had the graphics package Kaleidagraph fit the data, and the fit was excellent, with the equation:
$$x=0.18511t^{2.1572}$$
What does that give you for the work at t = 2 sec?

Chet

6. Dec 8, 2015

dsb1713

This equation represents x as a function of t, taken from the data provided in the problem?
I'm not certain how could I use this information to help me solve this problem

7. Dec 8, 2015

Staff: Mentor

Hmm. Since we'd expect $d = \frac{1}{2} a t^2$ from the standard kinematics, why not fit 2d vs t2?

8. Dec 8, 2015

Staff: Mentor

Yes.
Do you know how to determine the velocity from this equation?

9. Dec 8, 2015

Staff: Mentor

It wasn't clear to me from the problem statement that the force F was constant. If it is, then, of course, the fit you recommend is preferred.

OOPs. It did say constant. Oh well, Neeevvveeerr mind.

Chet

10. Dec 8, 2015

dsb1713

I assume through derivatives, although this is not in the scope of my current course. Also, yes, I believe the acceleration and force are both constant.
Thanks for your time & help