Question

A 3.0-kg block is on a horizontal surface. The block is at rest when, at t = 0, a force (magnitude P = 12 N) acting parallel to the surface is applied to the block causing it to accelerate. The coefficient of kinetic friction between the block and the surface is 0.20. At what rate is the force P doing work on the block at t = 2.0 s?

Answer 1

Step-by-step explanation:

It is given that,

Mass of the block, m = 3 kg

Initially, the block is at rest, u = 0

Force acting on the block, P = 12 N

The coefficient of kinetic friction between the block and the surface is,
`\mu_k=0.2`

We need to find the rate is the force P doing work on the block at t = 2.0 s. The rate at which work is done is called the power. Let is equal to P'

Frictional force acting on the block,
`f=\mu_k mg`

`f=0.2* 3\ kg* 9.8\ m/s^2=5.88\ N`

So, the net force acting on the block, F = P - f

`F=12-5.88=6.12\ N`

Let a is the acceleration of the block,
`a=(F)/(m)`

`a=(6.12)/(3)=2.04\ m/s^2`

Let v is the velocity of the block after 2 seconds. So,

`v=u+at`

`v=0+2.04* 2`

v = 4.08 m/s

Power,
`P'=(W)/(t)=(F.d)/(t)=F.v`

`P'=12\ N* 4.08\ m/s=48.96\ Watts`

So, the force P is doing work on the block at the rate of 48.96 watts.