Question

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

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Answer 1

Answer:

-4N

Explanation:

Refer to attachment please.

Answer 2

$\huge\star\underline\mathfrak\red{Answer:-}$

Given :

u = 20m/s

v= 0m/s

s = 50m

According to the third equation of motion,

${v}^{2} = {u}^{2} + 2as$

where,

Acceleration, a,

${0}^{2} = {20}^{2} + 2 \times a \times 50$

a = -4m/s^2

The negative sign indicates that acceleration is acting against the motion of the stone.

Mass of the stone, m = 1kg

From Newton's second law of motion,

Force,

F = mass × acceleration

F = ma

F = 1×(-4)

F = -4N

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

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