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Q.

A force F equals to 100 N is applied on 30 kg box inside which a 10 kg mass is hanging with the help of a massless string as shown in figure. Coefficient of friction between box and surface is 0.1. Initially system is at rest, then at this instant acceleration of box will be

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a

4 m/s2

b

6 m/s2

c

2 m/s2

d

1.5 m/s 2

answer is C.

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Detailed Solution

Fnet =F−f=100−40=60N⇒a=Fnet m=2m/s2
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A force F equals to 100 N is applied on 30 kg box inside which a 10 kg mass is hanging with the help of a massless string as shown in figure. Coefficient of friction between box and surface is 0.1. Initially system is at rest, then at this instant acceleration of box will be