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JEE Main 202311 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
11 April
Shift
Morning Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2023 · 11 April · Morning Shift · Physics · Work, Power and Energy

A force → F = (2 + 3x) ^ i acts on a particle in the x direction...

A force \overset{\to}{F} = \left(2 + 3x\right)\overset{^}{i} acts on a particle in the xx direction where FF is in Newton and xx is in meter. The work done by this force during a displacement from x=0x = 0 to x=4mx = 4m is _____ JJ.

Correct answer

0

Step-by-step solution

The formula to calculate the work done by a variable force is given by

W=∫x1x2Fdx...(1)W = {∫}_{{x}_{1}}^{{x}_{2}}Fdx . . . \left(1\right)

Substitute the values of the known parameters into equation (1) to calculate the required work done.

{W=∫04(2+3x)dx=[2x+32x2]04=[8+24]J=32J\begin{cases}W & = & {∫}_{0}^{4} ( 2 + 3x ) dx \\ & = & {\left[2x + \dfrac{3}{2}{x}^{2}\right]}_{0}^{4} \\ & = & [ 8 + 24 ] J \\ & = & 32J\end{cases}

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