The figure gives the acceleration of a 6.0 kg particle as an applied force moves

Question

The figure gives the acceleration of a 6.0 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 = 8.0 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 4.0 m/s², so the average force is F1 = ma = 6kg * 4.0m/s² = 24 N, so the work done over the first second is W1 = 24N * 1m = 24 J.

For the next three seconds, the force F3 = 6kg * 8 m/s² = 48 N
so the work done is W3 = 48 N * 3m = 144 J.

(a) work W4 = W1 + W3 = 168 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 - 8 m/s²,

so the average force F7 = - 48 N, and

the cumulative work done W7 = W4 + W6 = 120 J

(c) the incremental work from x=7m to x=9m is -72 J,

so the cumulative work done at x = 9 m is

Wnet = W7 + W9 = 48 J.

(d) At x = 4 m, KE = 168 J = ½ * 6kg * v² v = 7.48 m/s
(e) At x = 7 m, KE = 120 J = ½ * 6kg * v² v = 6.32 m/s
(f) At x = 9 m, KE = 48 J = ½ * 6kg * v² v = 4 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---