Question

A stone of 1 kg 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?​

Answer 1

Answer:

Explanation:

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?

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

Final velocity of the stone, v= 0

Distance covered by the stone, s= 50 m

We know the third equation of motion

v²=u²+2as

Substituting the known values in the above equation we get,

0² = (20)²+2(a)(50)

-400 = 100a

a = -400/100 = -4m/s² (retardation)

We know that

F = m×a

Substituting above obtained value of a=-4 in F= m x awe get,

F = 1×(-4) = -4N

(Here the negative sign indicates the opposing force which is Friction)

Answer 2

$\huge\bigstar\underline\mathfrak{Answer:-}$

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)

= -4N

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