A stone of mass 1 kg is thrown with a velocity 20 m/s 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 ice?
Answers
Answer:
Initial velocity of the stone, u= 20 m/s
Final velocity of the stone, v= 0
Distance covered by the stone, s= 50 m
We know the third equation of motion
v²=u²+2as
Substituting the known values in the above equation we get,
0² = (20)²+2(a)(50)
-400 = 100a
a = -400/100 = -4m/s² (retardation)
We know that
F = m×a
Substituting above obtained value of a=-4 in F= m x awe get,
F = 1×(-4) = -4N
(Here the negative sign indicates the opposing force which is Friction)
Answer:
Force, F = - 4 N
Explanation:
[Concept Used: Stopping distance and laws of motion]
Given;-
Mass, M = 1 kg
Initial Velocity, Vi = 20 m/s
Final Velocity, Vf = 0 m/s
Distance(here displacement), s = 50 m
Now, from third equation of motion we have;-
∵ Vf² - Vi² = 2as
∴ 0² - 20² = 2 × a × 50
a = - 400/100
a = - 4 m/s²
We also know that;-
∵ F = ma [where, f= force, m= mass, a = acceleration]
∴ F = 1 × (-4)
F = - 4 N
Hence, the force of friction between the stone and ice is - 4 N.
Hope it helps! ;-))