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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 traveling a distance of 50 m. What is the force of friction between the stone and the ice?

Updated On: Jan 19, 2026
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Solution and Explanation

Given: 

Mass of stone, \( m = 1 \, \text{kg} \)
Initial velocity, \( u = 20 \, \text{m/s} \)
Final velocity, \( v = 0 \, \text{m/s} \)
Distance traveled, \( s = 50 \, \text{m} \)

Step 1: Find acceleration

Using the equation of motion: \[ v^2 = u^2 + 2 a s \] \[ 0 = (20)^2 + 2 \cdot a \cdot 50 \] \[ 0 = 400 + 100 a \] \[ a = -4 \, \text{m/s²} \] (Negative sign indicates deceleration)

Step 2: Find frictional force

Using Newton's second law: \[ F = m \cdot a \] \[ F = 1 \times 4 = 4 \, \text{N} \]

Answer:

The force of friction between the stone and the ice = 4 N (opposite to the motion)

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Top Questions on Newton’s Second Law Of Motion

Questions Asked in CBSE Class IX exam

(Street Plan) : A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction.

All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines. There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North - South direction and another in the East - West direction. Each cross street is referred to in the following manner : If the 2nd street running in the North - South direction and 5th in the East - West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find: 

(i) how many cross - streets can be referred to as (4, 3). 

(ii) how many cross - streets can be referred to as (3, 4).

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