The figure gives the acceleration of a 4.0 kg particle as an applied force moves
ID: 1549981 • Letter: T
Question
The figure gives the acceleration of a 4.0 kg particle as an applied force moves it from rest along an x axis from x = 0 to x = 9.0 m. The scale of the figure's vertical axis is set by a_s = 9.0 m/s^2. How much work has the force done on the particle when the particle reaches (a) x = 4.0 m, (b) x = 7.0 m, and (c) x = 9.0 m? What is the particle's speed and direction (give positive answer if the particle moves along x axis in positive direction and negative otherwise) of travel when it reaches (d) x = 4.0 m, (e) x = 7.0 m, and (f) x = 9.0 m?Explanation / Answer
For the first second, the average acceleration is 4.5 m/s², so the average force is F1 = ma = 4kg * 4.5m/s² = 18 N, so the work done over the first second is W1 = 18N * 1m = 18 J.
For the next three seconds, the force F3 = 4kg * 9m/s² = 36 N
so the work done is W3 = 36N * 3m = 108 J.
(a) work W4 = W1 + W3 = 126 J
(b) Note that the average acceleration from x = 4m to x = 6m is zero. Then the average force over that range is zero. For the next meter, the acceleration is -9 m/s², so the average force F7 = -36 N, and the cumulative work done W7 = W4 + W6 = 90 J
(c) the incremental work from x=7m to x=9m is -54 J, so the cumulative work done at x = 9 m is Wnet = W7 + W9 = 36 J.
(This looks sound to me since the area above the graph is one "column" greater than the area below the graph, and each full column is 4kg * 9m/s² = 36 J of work.
(d) At x = 4 m, KE = 126 J = ½ * 4kg * v² v = 7.9 m/s
(e) At x = 7 m, KE = 90 J = ½ * 4kg * v² v = 6.7 m/s
(f) At x = 9 m, KE = 36 J = ½ * 4kg * v² v = 4.2 m/s