Question

A 2.0 kg wooden block is slid along a concrete surface (μk = 0.21) with an initial speed of 15 m/s. How far will the block slide until it stops?

Answer 1

The distance is 54.6 m

Explanation:

Given that,

Mass = 2.0 kg

Frictional coefficient = 0.21

Initial velocity = 15 m/s

We need to calculate the acceleration

Using formula of frictional force

`F = \mu mg`

`F=0.21*2.0*9.8`

`F = 4.12\ N`

We need to calculate the acceleration

`F = ma`

`a = (F)/(m)`

`a =(4.12)/(2.0)`

`a=2.06\ m/s^2`

We need to calculate the initial velocity

Using equation of motion

`v^2=u^2-2as`

Put the value

`0=15^2-2*2.06* s`

`s = (15^2)/(2*2.06)`

`s=54.6\ m`

Hence, The distance will be 54.6 m.