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Q.

A stone weighing 1 kg and sliding on ice with a velocity of 4 m/s is stopped by friction in 30 sec. The force of friction (assuming it to be constant) will be

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a

−0.13N

b

−0.2 N

c

0.2 N

d

20 N

answer is A.

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Detailed Solution

u=2 m/s, v=0, t=10 seca=v−ut=0−430=−430=−215=−0.13 m/s2∴Fraction force=ma=1×(−0.13)=−0.13 N
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