An object with a mass of 8 kg is moving at 7 m/s over a surface with a kinetic friction coefficient of 6 . How much power will it take to accelerate the object at #8 m/s^2?

1 Answer
Nov 25, 2017

P(v) = \mu_k.mg.v
\Delta P = P_f - P_i = \mu_k.mg.(v_f - v_i)
v_i = 7ms^{-1}; \qquad v_f = 8 ms^{-1}; \qquad \mu_k = 6; \qquad m = 8kg
\Delta P = 470 W.

Explanation:

As the object moves on a horizontal surface with friction, it looses kinetic energy. Friction converts its kinetic energy to thermal energy and dissipates it out. So to keep the object moving at a constant speed, external energy must be imparted at the same rate as its kinetic energy is dissipated out by friction.
First let us arrive at a relation connecting the power P, required to keep an object moving at a constant speed v.

Power: Power is defined as the rate at which energy is delivered-to or drawn-from a system. Here we are concerned with the mechanical energy, which is the sum of potential and kinetic energies :
dE = dU + dK

Since the object is moving on a horizontal surface there is no change in potential energy (dU = 0). Therefore, dE = dK.

By the Work-Energy Theorem, total work done is equal to the change in kinetic energy: dW = dK.

P \equiv (dE)/(dt) = (dK)/(dt) = (dW)/(dt)

The force doing the work here is the frictional force:
N=mg; \qquad F_f = \mu_k.N = \mu_kmg

Work done dW, by frictional force for a displacement dx is
dW = F_f.dx = \mu_kmgdx

Therefore, P = (dW)/(dt) = \mu_kmg\frac{dx}[dt} = \mu_kmg.v

This expression gives the power required to keep an object of mass m moving at a constant speed v, on a surface with a coefficient of kinetic friction \mu_k.

Given:
v_i = 7 ms^{-1}; \qquad v_f = 8 ms^{-1}; \qquad m = 8 kg; \qquad \mu_k = 6

P_i = \mu_kmgv_i = 3293 W; \qquad P_f = \mu_kmgv_f = 3763 W

So, to accelerate the body from a speed of 7 ms^{-1} to 8 ms^{-1} we need to impart an additional power of \Delta P = P_f - P_i = 470 W