Question

A block is shot with an initial velocity 5ms on a rough horizontal plane. Find the distance covered by
the block till it comes to rest. The coefficient of kinetic friction between the block and plane is 0.1.​

Answer 1

Answer:

Let m=mass of the block

V=5ms^-1. (v^2=25)

1/2mv^2=m/2=12.5

Answer 2

Answer:

A block is shot with an initial velocity of 5m/s on a rough horizontal plane. The coefficient of kinetic friction between the block and plane is 0.1.​ The distance covered by  the block till it comes to rest is 12.5 m.

Explanation:

Given,

Initial velocity (u)       = 5 m/s

Final velocity (v)        = 0 m/s

Coefficient of kinetic friction (μ)  =  0.1

we know,

$a=-\mu g$  ...(1)

where a is the retardation and g is the acceleration due to gravity.

g = 10 $\ m/s^{2}$

Substituting values to the equation (1),

a = $-0.1\times 10$ = -1 $m/s^{2}$

By using the equation of motion we can calculate the distance (s),

$\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{aS}$

$s = \frac{v^{2}- u^{2} }{2a}$

 $s = \frac{0^{2}- 5^{2} }{2\times-1} = \frac{25}{2} = 12.5 \ m$

ANS :

The distance covered by  the block = 12.5 m