Laws of Motion 3 Question 22

23. Masses M1,M2 and M3 are connected by strings of negligible mass which passes over massless and frictionless pulleys P1 and P2 as shown in 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 4.0kg 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.

If the mass M1 moves downwards with a uniform velocity, find

(1981, 6M)

(a) the mass of M1,

(b) the tension in the horizontal portion of the string.

(Take g=9.8m/s2,sin373/5 )

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Answer:

Correct Answer: 23. (a)

(a) 4.2kg, (b) 9.8N

Solution:

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

(a)

M1g=M2gsin37+μM2gcos37+μM3g or M1=M2sin37+μM2cos37+μM3

Substituting the values

M1=(4)35+(0.25)(4)45+(0.25)(4)=4.2kg

(b) Since, M3 is moving with uniform velocity

T=μM3g=(0.25)(4)(9.8)=9.8N



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