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

Answer:

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  

$F = \mu mg$

$F=0.21\times2.0\times9.8$

$F = 4.12\ N$

We need to calculate the acceleration

$F = ma$

$a = \dfrac{F}{m}$

$a =\dfrac{4.12}{2.0}$

$a=2.06\ m/s^2$

We need to calculate the initial velocity

Using equation of motion

$v^2=u^2-2as$

Put the value  

$0=15^2-2\times2.06\times s$

$s = \dfrac{15^2}{2\times2.06}$

$s=54.6\ m$

Hence, The distance will be 54.6 m.