. 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?

 

Best Answer

Sol. Given,

Initial velocity, u = 20 m/s

Final velocity, v = 0 (As the stone stops)

Acceleration, a =?

Distance travelled, s = 50 m

We know that,

v2= u2+2as

(0)2 = (20)2 + 2 × a × 50

 0 = 400 + 100 a

100 a = - 400

a = -400/100

a = - 4 m/s2

As,

Force, F = m × a

 F = 1 × (-4)

 F = - 4 N

Therefore, the value of force is – 4 N. Here, the negative sign indicates that the stone opposes the motion of stone.

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