Question

 A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes into rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice ​

Answer 1

Answer:

Mass = 1kg

u = 20m/s

v = 0

s = 50m

a= ?

V² - U² = 2as

=> 0² - 20² = 2 ( a ) ( 50 )

=> -400 = 100 a

=> a = - 400 / 100 = -4 m/s²

Therefore Frictional Force = Mass * Acceleration

=> Force = 1 kg * -4 m/s² = -4 kg.m/s² or -4 N

Hence the frictional force is -4 N.

Hope my answer helped !

Answer 2

꧁Ⓐ︎Ⓝ︎Ⓢ︎Ⓦ︎Ⓔ︎Ⓡ︎꧂

Given :

Initial velocity of the stone, u = 20 m/s.

Final velocity of the stone, v = 0 m/s.

Mass of the stone, m = 1 kg.

Distance covered by the stone, s = 50 m.

Explanation :

By third equation of motion,

v² - u² = 2as

0² - 20² = 2a × 50

- 400 = 100 a

-400/100 = a

-4 = a

➪ a = - 4 m/s²

The acceleration is -4 m/s²

F = ma

= 1 × -4

= - 4 N

Thus, force of friction between the stone and ice is - 4 N