Laws of Motion 3 Question 1

1. Two blocks A and B of masses mA=1kg and mB=3kg are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2 . The maximum force F that can be applied on B

horizontally, so that the block A does not slide over the block B is

[Take, g=10m/s2 ]

(2019 Main, 10 April II)

(a) 12N

(b) 16N

(c) 8N

(d) 40N

Objective Question II (One or more correct option)

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Answer:

Correct Answer: 1. (b)

Solution:

  1. Acceleration a of system of blocks A and B is

a= Net force  Total mass =Ff1mA+mB

where, f1= friction between B and the surface

=μ(mA+mB)g

So,

a=Fμ(mA+mB)g(mA+mB)

Here, μ=0.2,mA=1kg,mB=3kg,g=10ms2

Substituting the above values in Eq. (i), we have

a=F0.2(1+3)×101+3a=F84

Due to acceleration of block B, a pseudo force F acts on A.

This force F is given by F=mAa

where, a is acceleration of A and B caused by net force acting on B.

For A to slide over B; pseudo force on A, i.e. F must be greater than friction between A and B.

mAaf2

We consider limiting case,

mAa=f2mAa=μ(mA)g=μg=0.2×10=2ms2

Putting the value of a from Eq. (iii) into Eq. (ii), we get

F84=2F=16N



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