Question

An object with a mass of `8 kg` is moving at `27 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

2998W

Explanation:

While moving with uniform velocity overcoming the resisting force like friction ,the force equal in magnitude of the frictional force has to be applied on the object to continue its motion with uniform velocity.

This frictional force `F_"friction"=mu_kxxN=mu_kmxxg`
Where
`mu_k ="coefficient of kinetc friction"=0.6`
(it should be less than one)

`N = "Normal reaction"=mxxg, " "m= "mass of the object"`
`g= "Acceleration due to gravity"=9.8ms^-2`

When the body moves with acceleration (`a`) then the total force acting on the body at any instant will be sum of the resisting force and the force that produces acceleration

So Total force

`F_"total"=F_"friction"+F_"acceleration"=mu_kmg+ma...(1)`

The power required for this purpose will be the product of the resisting force and its velocity .

So

• `P =F_"total"xxV......(2)`

Where:
- `P="Power at any instant"`

• `F_"total"="Force on the body at that instant"`

• `V="Velocity of the body at that instant"`

Inserting the value of `F_"total"` in equation (2) we get

`P =(mu_kmg+ma)xxV`

This relation shows that Power is a function of Velocity (V) only when mass of the object, acceleration applied and coefficient of friction remain unaltered
As the body is moving with a constant acceleration we can have the power at the instant, it starts accelerating with initial velocity `V=27m/s`.

By the problem we have
`m=8kg,mu_k=0.6,V=27m/s,a=8ms^-2`

Hence

Power `P=(0.6xx8xx9.8+8xx8)xx27W=2998W`