A stone of 1 kg is thrown with a velocity of 20ms^-1 across a 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?
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10
Given :-
Mass of stone = m = 1 Kg
Velocity of stone = u = 20 ms-¹
Distance travelled = S = 50 m
Using third equation of Motion :-
v² = u² - 2aS
As Final velocity of stone is zero.
u² = 2aS
(20)² = 2(50)a
400 = 100a
a = 4 ms-²
Force acting on the body will be equal to the frictional force of the body.
F = f = ma
f = 1(4)
f = 4N.
Hence,
Frictional force = f = 4N
It will be acting in the opposite direction of Motion.
Answered by
4
Answer:
Given :-
Mass of stone = m = 1 Kg
Velocity of stone = u = 20 ms-¹
Distance travelled = S = 50 m
Using third equation of Motion :-
v² = u² - 2aS
As Final velocity of stone is zero.
u² = 2aS
(20)² = 2(50)a
400 = 100a
a = 4 ms-²
Force acting on the body will be equal to the frictional force of the body.
F = f = ma
f = 1(4)
f = 4N.
Hence,
Frictional force = f = 4N
- It will be acting in the opposite direction of Motion.
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