# 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?

Nov 25, 2017

$P \left(v\right) = \setminus {\mu}_{k} . m g . v$
$\setminus \Delta P = {P}_{f} - {P}_{i} = \setminus {\mu}_{k} . m g . \left({v}_{f} - {v}_{i}\right)$
v_i = 7ms^{-1}; \qquad v_f = 8 ms^{-1}; \qquad \mu_k = 6; \qquad m = 8kg
$\setminus \Delta P = 470$ $W$.

#### 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 :
$\mathrm{dE} = \mathrm{dU} + \mathrm{dK}$

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

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

$P \setminus \equiv \frac{\mathrm{dE}}{\mathrm{dt}} = \frac{\mathrm{dK}}{\mathrm{dt}} = \frac{\mathrm{dW}}{\mathrm{dt}}$

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

Work done $\mathrm{dW}$, by frictional force for a displacement $\mathrm{dx}$ is
$\mathrm{dW} = {F}_{f} . \mathrm{dx} = \setminus {\mu}_{k} m g \mathrm{dx}$

Therefore, $P = \frac{\mathrm{dW}}{\mathrm{dt}} = \setminus {\mu}_{k} m g \setminus \frac{\mathrm{dx}}{\mathrm{dt}} = \setminus {\mu}_{k} m g . v$

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 $\setminus {\mu}_{k}$.

Given:
${v}_{i} = 7$ ms^{-1}; \qquad v_f = 8 ms^{-1}; \qquad m = 8 kg; \qquad \mu_k = 6

${P}_{i} = \setminus {\mu}_{k} m g {v}_{i} = 3293$ W; \qquad P_f = \mu_kmgv_f = 3763# $W$

So, to accelerate the body from a speed of $7$ $m {s}^{- 1}$ to $8$ $m {s}^{- 1}$ we need to impart an additional power of $\setminus \Delta P = {P}_{f} - {P}_{i} = 470$ $W$