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A stone of 1 kg is thrown with a velocity of 20 m s⁻¹ acr... - Science
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?
Given Data,
Mass (
m):1 \, \text{kg}Initial velocity (
u):20 \, \text{m/s}Final velocity (
v):0 \, \text{m/s}(since the stone comes to rest)Distance (
s):50 \, \text{m}
Finding Acceleration (a)
Since we know the velocities and distance but not the time, we use the third equation of motion:
v^2 - u^2 = 2as
Substituting the values:
(0)^2 - (20)^2 = 2 \times a \times 500 - 400 = 100a-400 = 100a
Now, solve for a:
a = \frac{-400}{100}a = -4 \, \text{m/s}^2
(The negative sign indicates retardation, meaning the stone is slowing down due to friction.)
Finding the Force of Friction (F)
According to Newton's Second Law of Motion:
F = m \times a
Substituting the values:
F = 1 \, \text{kg} \times -4 \, \text{m/s}^2F = -4 \, \text{N}
The negative sign indicates that the force is acting in the direction opposite to the motion of the stone.
The magnitude of the force of friction between the stone and the ice is 4 \, \text{N}.