Question

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?

Answer 1

`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`