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Arrangement of two block system is as shown in the figure . mass of block A is m _A = 5 Kg and mass of B is m _B = 10 Kg A constant force F = 100 N applied on upper block A . Friction between A and B is \mu . and b/w B and ground surface is smooth , then find work done by frictional force on block B in t = 2 sec if system starts from rest A. 50 J B. 100 J C. 125 J D. 150 J

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@Alroy fernandes 

here is the complete question

Arrangement of two block system is as shown in the figure . mass of block A is m _A = 5 Kg 

and mass of B is m _B = 10 Kg 

A constant force F = 100 N applied on upper block A . Friction between A and B is \mu . and b/w B and ground surface is smooth , then find work done by frictional force on block B in t = 2 sec  if system starts from rest 

 For block A 

N = mg    f_K = \mu N

=  = \mu mg = \mu M_n g = 0.5 \times 5 \times 10 = 25 N

For block B 

f_K = m_B a = \mu m_A g \\ a = \frac{\mu m_A g }{m_B } =

=\frac{ 0.5 \times 5 \times 10 }{10}= 2.5 m/s^2

s = \mu t + 1/2 at^2 \Rightarrow s = 1/2 \times 2.5 \times 4 = 5 m

W = F_k \times s \cos 0 = 25 \times 5 \times 1 = 125 J

 

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Safeer PP

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