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Question

A mass of 2 kg is placed on a trolley of 20 kg sliding on a smooth surface. The coefficient of friction between the mass and the surface of the trolley is 0.25. A horizontal force of 2 N is applied to the mass. The acceleration of the system and the frictional force between the mass and surface of trolley are (Take g=10 m/s2)


A
1.8 ms2, 0.09 N
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B
0.9 ms2, 18 N
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C
0.09 ms2, 1.8 N
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D
1 ms2, 2 N
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Solution

The correct option is C 0.09 ms2, 1.8 N
The FBDs of the system is shown as


From the FBD of block,
N1=2g=20 N
From the FBD of trolley,
N2=N1+20g=20+200=220 N

Limiting frictional force fmax=μN1=0.25×20=5 N
fmax> applied external force of 2 N, the block and trolley will not have relative motion and they will move together.

Let the common acceleration be ac
Then, 2=(20+2)ac
ac=222=111=0.09 ms2

Frictional force is given from the FBD of trolley as
f=20ac=20×0.09=1.8 N

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