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+2 votes
62.2k views
in Physics by (53.3k points)
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Three masses M = 100 kg, m1 = 10 kg and m= 20 kg are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force F is applied on the system so that the mass m2 moves upward with an acceleration of 2 ms-2. The value of F is: 

(Take g = 10 ms-2)

(A) 3360 N

(B) 3380 N

(C) 3120 N

(D) 3240 N

2 Answers

+3 votes
by (15.0k points)
selected by
 
Best answer

Correct option is (A) 3360 N

If T is the tension in thread and a be acceleration of 100kg block

For the 10 kg block kept on the 100 kg block (In the 100 kg block frame of reference,

10a - T = 10 x 2  ......(1)

For the 20 kg block in vertical direction,

T - 20g = 20 x 2

⇒ T - 20 x 10 = 20 x 2    .......(2)

Adding the above two equations,

⇒ 10a = 3 x 20 + 20 x 10 = 260

a = 26ms-2, T = 240 N

The applied force F will be responsible for the horizontal acceleration of 100 kg and 20 kg block.

Therefore,

F - T = 120 a

⇒ F = 3360 N

+5 votes
by (53.5k points)

Correct option is (A) 3360 N

Let acceleration of 100 kg block = a1

FBD of 100 kg block w.r.t ground

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