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dsb1713

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## Homework Statement

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.

## Homework 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

## 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?

Thanks for your time.