A stone of mass 1 kg thrown with an initial velocity of 20 m/s over a frozen lake comes to rest after 10 s . Find the magnitude of the force.
a)2 N
b)1 N
c)-2 kgm/s2
d)-1 kgm/s2
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Answer:
-4N
Explanation:
Given:
The initial velocity of the stone, u= 20 m/s
The final velocity of the stone, v= 0
Distance covered by the stone, s= 50 m
Find out:
The force of friction between the stone and the ice
Solution
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 a we get,
F = 1 × (-4) = -4N
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