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}{20}=1.25m/s^{2} $
$ For6kg:-F-2T=6a…( i ) $
$ For2kg:-T-2g=2(2a)…( ii ) $ From (i) and (ii), $ F=75N $
BETA
