NEET Solved Paper 2019 Question 15
Question: A force $ F=20+10y $ acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from $ y=0toy=1 $ m is: [NEET 5-5-2019]
Options:
A) 25 J
B) 20 J
C) 30 J
D) 5 J
Show Answer
Answer:
Correct Answer: A
Solution:
$ F=20+10y $
$ W=\int\limits_0^{1}{(2010y)dy} $
$ =20[y]_0^{1}+[ \frac{10y^{2}}{2} ]_0^{-1} $
$ =20+5 $
$ =25 $
