Work Energy And Power Question 216
Question: The position of a particle of mass 4 g, acted upon by a constant force is given by $ x = 4t^{2} + t $ , where x is in metre and t in second. The work done during the first 2 seconds is
Options:
A) $ 128mJ $
B) $ 512mJ $
C) $ 576mJ $
D) $ 144mJ $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] here, $ m=4,g=4\times 1{0^{-3}}kg $
$ x=4t^{2}+t\therefore \frac{dx}{dt}=8t+1\frac{d^{2}x}{dt^{2}}=8 $
Work done. $ W=\int{fdx}=\int{m\frac{d^{2}x}{dt^{2}}( \frac{dt}{dt} )dt} $
$ =\int\limits_0^{2}{( 4\times {10^{-3}} )( 8 )( 8t+1 )dt} $
$ =32\times {10^{-3}}\int\limits_0^{2}{( 8t+1 )dt =32\times 1{0^{-3}}{{[ \frac{8t^{2}}{2}+t ]}^{2}}_0} $
$ =32 \times 1{0^{-3}} [ 4{{( 2 )}^{2}}+2-0 ]=576mJ $
BETA
