Question

Power: A block of mass m = 4.0kg is at rest on a horizontal frictionless floor. At time t = 0 a horizontal force F = Fyi is applied to the block. At an instant of t later the block is at x = 2.0t+ (x is in meters) from the block's initial position, x = 0. Part a. Find Fx, the x-component of the force F as function of time. (3 pts) Ex(t) = Part b. Find the instantaneous power created by force F as function of time. (3 pts) P(t) =

Answer 1

For the first part use s = u * t + 1 / 2 * a * t^2. Where s is the distance covered in time t which is 2 * t^4. And u is the initial velocity of the particle, and here u = 0, since the particle starts from rest. So 2 * t^4 = 1 / 2 a * t^2. a is the acceleration, which is a = 4 * t^2 m / s^2. Now F = m * a = 4 * 4 * t^2 N = 16 * t^2 N = (4 * t)^2 N. Here mass, m = 4.0 kg.

Now Power, P = work done per unit time. So work done W = Force * displacement = (4 * t)^2 * 2 * t^4 J = 32 * t^6. So P = 32 * t^6 / t W = 32 * t^5 W.