Work Energy and Power | Physics JEE previous year questions with Solutions

0

072920 77839
info@smartachievers.in

Topic Question Set

Q 1 :
A force $(3 x^{2} + 2 x - 5) N$ displaces a body from $x = 2 m$ to $x = 4 m .$ Work done by this force is __________ J. [2024]

0%

(58) $Force, F = (3 x^{2} + 2 x - 5) N, x_{1} = 2 m$ $a n d$ $x_{2} = 4 m$

$Work done, W = \int_{x_{1}}^{x_{2}} F \cdot d x = \int_{2}^{4} (3 x^{2} + 2 x - 5) d x$

$W = {[x^{3}]}_{2}^{4} + {[x^{2}]}_{2}^{4} - 5 {[x]}_{2}^{4} = [64 - 8] + [16 - 4] - 5 [4 - 2] = 58 J$

Want Us to Email you About Special Offers & Updates?