50 kg uniform crate initially rests on the inclined plane If

50 kg uniform crate initially rests on the inclined plane. If a force (F= 40t + 200)as shown starts to pull the crate, determine the power output developed by the force at t = 6 sec. The coefficients of static and kinematic friction between the crate and the surface are Mu_s = 0.2 and Mu_k = 0.15. Units of force are newtons and time are seconds.

Solution

We have a time varying force acting on a block. We know that for a force acting on a body, say F, and moving the body at velocity, say V, the power output is given as F X V. So as to determine the power output at t = 6 s, we will need to find the value of F at t =6 which can be directly obtained from the given equation. Also, we need to determine the velocity of the block at the given time for which we will use the net force acting on the block and then use the equation dv/dt = acceleration to determine the velocity at t =6 seconds.

The block of mass m = 50 kg rests on an incline plane of 30 degrees and is subject to a force of 40t + 200.

Also the kinetic and static coefficients of friction are given as: Uk = 0.15 and Us = 0.2

Now as the block rests on the incline the net friction force acting on it would be: MgCos30 x Us

or, Friction = 50 x 9.81 x Cos30 x 0.2 = 84.9571 N

That would mean the block will start off the moment we start to apply the force.

Now, acceleration = 40t + 200 - 50 x 9.81 x Cos30 x 0.15 =

(0.8t + 4 - 1.2744) m/s^2

or dv/dt = (0.8t + 2.7256)

or, dv = (0.8t + 2.7256)dt

Integrating the above expression from t = 0 to t =6 we get:

V = 0.8 x 18 + 2.7256(6) = 30.7536 m/s

Now the force acting on the block at t = 6 is given as: F = 40(6) + 200 = 440 N

Therefore the required power output = F x V = 440 x 30.7536 = 13531.584 Watts