# A Stone of 1kg is Thrown With a Velocity of 20ms-1 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?

By BYJU'S Exam Prep

Updated on: September 25th, 2023

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

Step 1: Given that Mass of stone, m = 1kg

The final velocity of the stone, v= 0

The initial velocity of the stone, u= 20ms-1

Distance covered by the stone, s = 50m

Step 2: Now we have to use below mentioned formulae

v2=u2+2as

F=m×a, where F is force.

Step 3: Now we have to calculate the acceleration and force of friction

Using,

v2=u2+2as

Substituting the formula in above formula we get

0=202+2×a×50

On rearranging we get

100a=-400

a=-400/100

a=-4ms-2

Hence, the acceleration of stone is -4ms-2. Here, -ve sign shows retardation.

We have

F=m×a

Substituting the values in the above formula

F=m×a

F=1×(-4)

F=-4N

The -ve symbol here indicates that the force, which is the frictional force, is acting in the opposite direction.

### Factors affecting Frictional Force

1. Surface roughness and the amount of force pressing them together have the biggest impact on these forces.
2. The amount of frictional force is influenced by the object’s position and angle.
3. The frictional force will be equal to the weight of the object if it is put flat against another object.
4. The frictional force will increase and surpass the weight of the thing if an object is pushed up against the surface.

Summary:-

## A Stone of 1kg is Thrown With a Velocity of 20ms-1 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?

A stone of 1kg is thrown with a velocity of 20ms-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50m. The force of friction between the stone and the ice is -4 N.

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