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Q6:
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?

Solution :

Step 1: Identification of Given Physical Quantities

To determine the force of friction acting on the stone, we identify the following variables provided in the problem statement:

  • Mass of the stone ($m$) = $1\text{ kg}$
  • Initial velocity ($u$) = $20\text{ m s}^{-1}$
  • Final velocity ($v$) = $0\text{ m s}^{-1}$ (as the stone comes to rest)
  • Distance traveled ($s$) = $50\text{ m}$

Step 2: Calculation of Acceleration

Using the third equation of motion, which relates velocity, acceleration, and displacement independent of time [Per Kinematic Laws of Motion]:

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

Substituting the known values:

$0^2 = (20)^2 + 2 \cdot a \cdot 50$

$0 = 400 + 100a$

$-400 = 100a$

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

The negative sign indicates that the acceleration is directed opposite to the motion, confirming the action of a retarding force (friction).

Step 3: Calculation of Frictional Force

According to Newton’s Second Law of Motion, the force acting on an object is the product of its mass and acceleration ($F = ma$) [Per Newton's Second Law]:

$F = 1\text{ kg} \times (-4\text{ m s}^{-2})$

$F = -4\text{ N}$

Visual Representation of Forces

Stone Motion ($v$) Friction ($F$)

Analytical Summary

Parameter Value
Mass $1\text{ kg}$
Deceleration $4\text{ m s}^{-2}$
Frictional Force $4\text{ N}$ (Opposing motion)

Final Solution: The force of friction between the stone and the ice is 4 N in the direction opposite to the motion of the stone.


More Questions from Class 9 Science Force and Laws of Motion Exercises


CBSE Solutions for Class 9 Science Force and Laws of Motion


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