Laws Of Motion Question 136

Question: The system starts from rest and $ A $ attains a velocity of 5 m/s after it has moved 5 m towards right. Assuming the arrangement to be frictionless everywhere and pulley and strings to be light, the value of the constant force $ F $ applied on A is

Options:

A) 50 N

B) 75 N

C) 100 N

D) 96 N

Show Answer

Answer:

Correct Answer: B

Solution:

[b] $ a=\frac{v^{2}}{2s}=\frac{25}{10}=2.5m/s^{2} $

$ For6kg:-F-2T=6a…( i ) $

$ For2kg:-T-2g=2(2a)…( ii ) $ From (i) and (ii), $ F=75N $



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