Get Answers to all your Questions

header-bg qa

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 . Coefficient of friction between A and B is \mu  and b/w B and ground surface is zero , then find work done (in Joule) by frictional force on block B in t = 2 sec  if system starts from rest.

Option: 1 50

Option: 2 100

Option: 3 125

Option: 4 150

Answers (1)

best_answer

As we have learned

Work done by the frictional force is positive -

When force is applied on a body, which is placed above another body ,the work done by the frictional force on the lower body may be positive 

-

 

for block A,

N = mg 

  f_K = \mu N

  \Rightarrow f_{k}= \mu m_{A}g = 0.5 \times 5 \times 10 = 25 N

For block B 

f_K = m_B a

\Rightarrow \mu m_A g =m_{B}a

\Rightarrow 25=10\times a\Rightarrow a=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

 

 

Posted by

Gautam harsolia

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks