F_E = mg g = 9.80 m/s^2 = 35.2 ft/s^2 W = bar F middot delta bar r = F delta r c
ID: 1549939 • Letter: F
Question
F_E = mg g = 9.80 m/s^2 = 35.2 ft/s^2 W = bar F middot delta bar r = F delta r cos theta W = integral^_ F(x) dx K = 1/2 mv^2 W_ = K_r - K_i A 58.8 N box is subject to a net force given by F(x) = 200 N -(3 N/m^3) x^3 The box has an initial speed of 7.0 m/s at x = 2.0 m, and moves in the + x direction, to final position x = 40 m. a) What is the work done by this non-constant net force? b) What is the box's initial kinetio energy, at x = 2.0 m? c) What is the box's final kinetic energy, at x = 4.0 m? d) What is the box's speed when it is at x = 4.0 m?Explanation / Answer
work done = F.dx
work done = (200 - 3x^3) dx
integrating the equation we'll get
W = 200x - 3 * x^4 / 4
work done = (200 * 4 - 3 * 4^4 / 4) - (200 * 2 - 3 * 2^4 / 4)
work done = 220 J
initial kinetic energy = 0.5 * m * initial velocity^2
initial kinetic energy = 0.5 * 58.8 * 7^2
initial kinetic energy = 1440.6 J
work done = change in kinetic energy
work done = final kinetic energy - initial kinetic energy
220 = final kinetic energy - 1440.6
final kinetic energy = 1660.6 J
kinetic energy = 0.5 * mv^2
1660.6 = 0.5 * 58.8 * v^2
v = 7.515 m/s
speed at 4m = 7.515 m/s
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