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Q.

Two blocks of masses 10 kg and 5 kg are connected by a flexible but inextensible string over a shaft as shown in the figure. The system starts from rest with 10 kg mass moving downward. If a constant frictional force of 25 N acts at the shaft, the velocity when the 10 kg block has moved 2 m down is

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a

2.6ms−1

b

6.2ms−1

c

3.2ms−1

d

4ms−1

answer is A.

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Detailed Solution

a=10g−5g−2515=100−50−2515=53ms−2 Since, v2−02=2as⇒ v2=253(2)=203⇒ v=2.6ms−1
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