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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/$$s^2$$)
1032881_581dfc3b088c494285b6214f6a6bc5ed.PNG


A
0.5 m/s2
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B
0.4 m/s2
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C
1 m/s2
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D
56 m/s2
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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 force
so 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.


959962_1032881_ans_693e0538e726487684296056dac0e402.jpeg

Physics

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