Laws of Motion 3 Question 23

24. Two blocks connected by a massless string slides down an inclined plane having an angle of inclination of 37. The masses of the two blocks are M1=4kg and M2=2kg respectively and the coefficients of friction of M1 and M2 with the inclined plane are 0.75 and 0.25 respectively. Assuming the string to be taut, find (a) the common acceleration of two masses and (b) the tension in the string. (sin37=0.6,cos37=0.8).( Take g=9.8m/s2)

(1979)

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Answer:

Correct Answer: 24. (a) a=1.3m/s2

(b) T=5.2N

Solution:

  1. Maximum force of friction between M1 and inclined plane

f1=μ1M1gcosθ=(0.75)(4)(9.8)(0.8)=23.52NM1gsinθ=(4)(9.8)(0.6)=23.52N=F1

Maximum force of friction between M2 and inclined plane

f2=μ2M2gcosθ=(0.25)(2)(9.8)(0.8)=3.92NM2gsinθ=(2)(9.8)(0.6)=11.76N=F2

Both the blocks will be moving downwards with same acceleration a. Different forces acting on two blocks are as shown in above figure.

Equation of motion of M1

 or T+F1f1=M1aT=4a

Equation of motion of M2

F2Tf2=M2a or 7.84T=2a

Solving Eqs. (i) and (ii), we get

mgsinθ=(2)(10)12=10N=F1

(a) Force required to move the block down the plane with constant velocity.

F1 will be acting downwards, while F2 upwards.

Since F2>F1, force required

F=F2F1=11.21N

(b) Force required to move the block up the plane with constant velocity.

F1 and F2 both will be acting downwards.

F=F1+F2=31.21N



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