Question

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

Answer 1

Given:-

• Mass of Stone = 1kg

• Initial Velocity of Stone = 20m/s

• Final Velocity = 0m/s

• Distance Travelled = 50m.

To Find:-

• Force acting between stone and the Ice.

Formulae used:-

• v² - u² = 2as

• F = ma

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• S = Distance
• F = Force
• m = Mass

Now,

First we have to find Acceleration

→ v² - u² = 2as

→ (0)² - (20)² = 2 × a × 50

→ -400 = 2 × a × 50.

→ -400 = 100a.

→ a = -400/100

→ a = -4m/s²

Hence, The Retardation of stone is -4m/s²

Therefore,

→ F = ma

→ F = 1 × -4

→ F = -4N.

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

Answer 2

Question:

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

Given that:

u (Initial velocity)= 20m/s

v (Final velocity) = 0m/s

s (Distance covered) = 50m.

We need to find acceleration first using third equation of motion,

$v {}^{2} - u {}^{2} = 2 \times a \times s$

Since, v^2 = 0

➝ – 400 = 2 × a × s

➝ 400 = 2 × a × 50

➝ a = – 400/100 = – 4m/s

Hence, acceleration = – 4m/s

${F_net}$ = m × a

= 1 × (–4)

= – 4N (opposing force)

Suppose the object is moving in the right direction then here minus sign indicates that the force of the object is in the opposite side of the motion of that object.

Therefore, force of friction between the stone and the ice is – 4N.