more physics.....?

Question

A 1.7 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. A stroboscopic graph of the position of the body as it slides to the right. 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 = 2.0 s? (HINT: Find the average acceleration over t = 0 to t = 2.0 s.)

the body is at 0.004 at 0.5s, 0.2 at 1s, 0.404 at 1.5s, and 0.8 at 2s

Answer 1

The average velocities during each interval would be (0.004 m) / (0.5 s) = 0.008 m until 0.5 s, (0.2 m - 0.004 m) / (0.5 s) = 0.392 m/s from 0.5 s to 1s, (0.404 m - 0.2 m) / (0.5 s) = 0.408 m/s from 1s to 1.5 s, and (0.8 m - 0.404 m) / (0.5 s) = 0.792 m/s from 1.5 s to 2s. If we assume each velocity is reached at the end of a 0.5 s period, we can perform the same operation to calculate the accelerations during each period: 0.016 m/s^2, 0.768 m/s^2, 0.032 m/s^2, and 0.768 m/s^2. Now, multiply the mass of 1.7 kg by each acceleration to get the applied force during each period: 0.0272 N, 1.31 N, 0.0544 N, and 1.31 N. Then, multiply each force by the distance traveled during each period to get the work done during each: (0.0272 N)*(0.004 m) = 0.000109 W, (1.31 N)*(0.2 m - 0.004 m) = 0.256 W, (0.0544 N)*(0.404 m - 0.2 m) = 0.0111 W, and (1.31 N)*(0.8 m - 0.404 m) = 0.517 W. The sum of this work is 0.784 W.

Offhand, I would have chosen a different method to solve this. I'd simply take the final velocity of 0.792 m/s and calculate the kinetic energy, 0.5*m*v^2 = 0.5*(1.7 kg)*(0.792 m/s)^2 = 0.533 W. The answers are different because you can't get truly accurate information from a finite number of discrete points on a curve.