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NCERT

Class 9

Science

Solutions

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?

According to question,

m = 1 kg, initial velocity u = 20 ms^{–1}

final velocity v = 0, distance s = 50 m

∴ from the formula

v^{2 }= u^{2 }+ 2as

0 = 20^{2} + (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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