Question

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

Answer 1

Answer:

mass = 1 kg

u = 20 m/s

v = 0 m/s

s = 50 m

v^2 - u^2 = 2 (as)

0- 400 = 2 (50) (a)

-400 = 100 a

-4 m/s^2 = a

force = mass × acceleration

= 1 × -4

= -4 N

Answer 2

Answer:

• -4N

Explanation:

Given

• The initial velocity of the stone, u = 20 ms-¹
• The final velocity of the stone, v = 0
• Distance covered by the stone before coming to rest s = 50 m
• Let a be the constant acceleration/retardation of the stone.

Tofind

• force of friction between the stone and the ice?

Solution

Using third eq of motion,

v² - u² = 2as

where,

• v = final velocity
• u = initial velocity
• a = acceleration
• s = displacement

⇒ 0²- 20² = 2 × a × 50

⇒ a = - ( 20²/2×50 )

⇒ a = -4ms-²

Using Newton's second law

F = ma

where

• F = force
• m = mass
• a = acceleration

⇒ F = 1 × (-4)

⇒ F = -4N

Hence

• Friction force acting between the stone and the ice is −4N