Question:
A stone of 1 kg is thrown with a velocity of 20 ms–1 across the frozen surface of lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Answer:
According to question,
m = 1 kg, initial velocity u = 20 ms–1
final velocity v = 0, distance s = 50 m
∴ from the formula
v2 = u2 + 2as
0 = 202 + (2 x a x 50)
0 = 400 + 100a
-400/100 = a
a = -4 ms–2
Frictional force F = ma
F = 1 X (-4)
F = -4N
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