Neet Solved Paper 2015 Question 7
Question: A block of mass 10 kg, moving in x-direction with a constant speed of $ 10,m{s^{-1}}, $ is subjected to a retarding force $ F=0.1,\times ,J/m $ during its travel from $ x=20,m $ m to 30 m. Its final KE will be
Options:
A) 475 J
B) 450 J
C) 275 J
D) 250 J
Show Answer
Answer:
Correct Answer: A
Solution:
From work-energy theorem, Work done = Change in KE
$ \Rightarrow \ W=K _{t}-K _{i} $
$ \Rightarrow \ K _{f}=W+K _{i}=\int _{x _1}^{x _2}{Fxdx}+\frac{1}{2}mv^{2} $
$ =\int _20^{30}{-0.1x,dx}+\frac{1}{2}\times 10\times 10^{2} $
$ =-0.1[ \frac{x^{2}}{2} ] _20^{30}+500 $ $ =-0.05[30^{2}-20^{2}]+500 $ $ =-0.05[900-400]+500 $
$ \Rightarrow \ K _{f}=-25+500=475J $
BETA
