Question

# A block of mass $$M_1 = 10 \ kg$$ is placed on a slab of mass $$M_2$$ =  30 Kg. The slab lies on a frictionless horizontal surface as shown in figure. The coefficient of static friction between the block and slab is $$\mu_s$$ = 0.25 and that of dynamic friction is $$\mu_k$$ = 0.12. A force F = 40 N acts on block $$M_1$$. The acceleration of the slab will be (g = 10 m/$$s^2$$)

A
0.5 m/s2
B
0.4 m/s2
C
1 m/s2
D
56 m/s2

Solution

## The correct option is B 0.4 $$m/s^2$$Static Friction between the block is $$\mu_{k} M_{1}g=.25\times100=25N$$static friction < applied forceso in this case kinetic friction will be act  $$f_{k}= .12\times 100N=12N$$FBD diagram of both the block as shown  for $$M_{1}$$ block which as acceleration $$a_{1}$$        $$F-f_{K}=10a_{1}$$                      eq(1)..          $$40-12=10a_{1}$$           a_{1}=2.8ms^{-2} for $$M_{2}$$ slab which as acceleration            $$f_{k}=30a_{2}$$                eq(2)..            $$12=30a_{2}$$          $$a_{2}=0.4ms^{-2}$$solving eq(1)and eq(2)          accleration of slab is $$0.4ms^{-2}$$Hence the B option is correct.Physics

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