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Question: Answered & Verified by Expert

Masses, M1, M2 and M3 are connected by strings of negligible mass which pass over massless and frictionless pulleys P1 and P2 as shown in the figure. The masses move such that the portion of the string between P1 and P2 is parallel to the inclined plane and the portion of the string between P2 and M3 is horizontal. The masses M2 and M3 are 4.0 kg each and the coefficient of kinetic friction between the masses and the surfaces is 0.25. The inclined plane makes an angle of 37° with the horizontal and the mass M1 moves downwards with a uniform velocity. Find the tension in the horizontal portion of the string.
(Take g=9.8 m s-2, sin 37° 35)

PhysicsLaws of MotionJEE Main
Solution:
1156 Upvotes Verified Answer
The correct answer is: 9.8

Constant velocity means the net acceleration of the system is zero. Or the net pulling force on the system is zero. While calculating the pulling force, tension forces are not taken into consideration. Therefore,

 M1g=M2g sin 37°+μM2g cos37°+μM3g

or M1=M2 sin 37°+μM2 cos 37°+μM3
 
Substituting the values
 
M1=(4)(35)+(0.25)(4)(45)+(0.25)(4)      =4.2 kg

Since, M3 is moving with uniform velocity

T=μM3g=(0.25) (4) (9.8)=9.8 N

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