A 2.8 kg body is at rest on a frictionless horizontal air track when a horizontal force F acting in the positive direction of an x axis along the track is applied to the body. The force 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 F between t = 0 and t = 1.0 s?
Find the distance it went:
distance= 1/2 (F/m)t^2
then work is the dot product of force and distance.
To find the work done on the body by the applied force between t = 0 and t = 1.0 s, we need to first calculate the distance covered by the body during this time interval.
The formula you provided, distance = 1/2 (F/m) t^2, is incorrect. This formula actually calculates the displacement, not the distance. However, since the body is at rest initially, the displacement and distance will be the same in this case.
To calculate the distance covered, we need to find the average velocity of the body during the time interval and then multiply it by the time. The average velocity can be found by dividing the displacement by the time interval.
Since the body is at rest initially, the displacement will be equal to the distance covered. Hence, the formula becomes:
distance = (average velocity) * (time interval)
The average velocity can be calculated using the formula:
average velocity = displacement / time interval
Given that the time interval is 1.0 s and the body is at rest initially, the average velocity will be zero. Therefore, the distance covered by the body will also be zero.
Therefore, if the body does not move during the time interval, no work is done by the applied force F.