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A stone of 1 kg is thrown with a velocity of 20 ms-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?

Laws of Motion

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Answer

Given,

Mass (m) = 1 kg

Initial velocity (u) = 20 ms-1

Final velocity (v) = 0

Distance travelled (s) = 50 m

According to the third equation of motion,

2as = v2 - u2

or

a = v2−u22s\dfrac{\text{v}^2 - \text{u}^2 }{2\text{s}}

Substituting we get,

a = 0−(20)22×50=−400100\dfrac{0 - (20)^2}{2 \times 50} = -\dfrac{400}{100} = -4 [retardation]

Now, as Force = mass x acceleration

Substituting we get,

F = 1 × (-4) = -4 N

Negative sign indicates the opposing force which is friction

Hence, force of friction between the stone and the ice = -4 N

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