JEE Main Solved Paper 2017 Question 26
Question: A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be: [JEE Main Solved Paper-2017]
Options:
A) 9 J
B) 18 J
C) 4.5 J
D) 22 J
Show Answer
Answer:
Correct Answer: C
Solution:
- $ F=6t=ma $
$ \Rightarrow $ $ a=6t $
$ \Rightarrow $ $ \frac{dv}{dt}=6t $ $ \int_0^{v}{dv}=\int_0^{1}{6t},dt $ $ v=(3t^{2})_0^{1}=3,m/s $ From work energy theorem $ W _{F}=\Delta K.E=\frac{1}{2}m( v^{2}-u^{2} ) $ $ =\frac{1}{2}(1)(9-0)=4.5,J $
