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

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• Initial velocity of the stone, u = 20 m/s
• Final velocity of the stone, v = 0 (finally the stone comes to rest)
• Distance covered by the stone, s = 50 m
• According to the third equation of motion:
• v2 = u2 + 2as

Where,

• Acceleration, a
• (0)2 = (20)2 + 2 × a × 50
• a = −4 m/s2
• The negative sign indicates that acceleration is acting against the motion of the stone.
• Mass of the stone, m = 1 kg

From Newton’s second law of motion:

• Force, F = Mass × Acceleration
• F = ma
• F = 1 × (− 4) = −4 N

Hence, the force of friction between the stone and the ice is −4 N.

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Answer 2

$\boxed{\sf \green{Answer}}$

F = -4N

$\boxed{\sf \green{Explanation}}$

Given:

• m = 1kg
• u = 20m/s
• v = 0
• S = 50m

To Find:

• Frictional force between the ice and stone

Solution:

$\boxed{\sf \blue{{v}^{2}-{u}^{2}= 2as}}$

$\mathsf{ {0}^{2}-{20}^{2}= 2(-a) (50)}$

$\mathsf{ -400= 100a}$

$\mathsf{ a = -4 m{s}^{-2}}$

Note: -ve of acceleration indicates Retradation

$\boxed{\sf \pink{ F = ma}}$

$\mathsf{ F = (1)(-4)}$

$\mathsf{ F =-4N }$

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