Question

A stone of 1 kg is thrown with a velocity of 20ms^-1 across a frozen surface of a lake and comes to rest after travelling a distance of 50m .What is the force of friction between the stone and the ice? ​

Answer 1

Given :-

Mass of stone = m = 1 Kg

Velocity of stone = u = 20 ms-¹

Distance travelled = S = 50 m

Using third equation of Motion :-

v² = u² - 2aS

As Final velocity of stone is zero.

u² = 2aS

(20)² = 2(50)a

400 = 100a

a = 4 ms-²

Force acting on the body will be equal to the frictional force of the body.

F = f = ma

f = 1(4)

f = 4N.

Hence,

Frictional force = f = 4N

It will be acting in the opposite direction of Motion.

Answer 2

Answer:

Given :-

Mass of stone = m = 1 Kg

Velocity of stone = u = 20 ms-¹

Distance travelled = S = 50 m

Using third equation of Motion :-

v² = u² - 2aS

As Final velocity of stone is zero.

u² = 2aS

(20)² = 2(50)a

400 = 100a

a = 4 ms-²

Force acting on the body will be equal to the frictional force of the body.

F = f = ma

f = 1(4)

f = 4N.

Hence,

Frictional force = f = 4N

• It will be acting in the opposite direction of Motion.