(B) The work done $dW$ in stretching the rubber band by an infinitesimal distance $dx$ is given by $dW = F dx$.Substituting the given force $F = ax +

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$\frac{aL^2}{2} + \frac{bL^3}{3}$

(B) The work done $dW$ in stretching the rubber band by an infinitesimal distance $dx$ is given by $dW = F dx$.Substituting the given force $F = ax + bx^2$,we get $dW = (ax + bx^2) dx$.To find the total work done $W$ in stretching the rubber band from $0$ to $L$,we integrate the expression:$W = \int_{0}^{L} (ax + bx^2) dx$$W = \int_{0}^{L} ax dx + \int_{0}^{L} bx^2 dx$$W = a \left[ \frac{x^2}{2} \right]_{0}^{L} + b \left[ \frac{x^3}{3} \right]_{0}^{L}$$W = \frac{aL^2}{2} + \frac{bL^3}{3}$

Class 11 Physics Work, Energy, Power and Collision Work Done by Variable Force and Force-Displacement Graph

Medium

When a rubber band is stretched by a distance $x$,it exerts a restoring force of magnitude $F = ax + bx^2$,where $a$ and $b$ are constants. The work done in stretching the unstretched rubber band by $L$ is

JEE MAIN 2014 All JEE MAIN

A
$\frac{1}{2}(aL^2 + bL^3)$
B
$\frac{aL^2}{2} + \frac{bL^3}{3}$
C
$\frac{1}{2}(\frac{aL^2}{2} + \frac{bL^3}{3})$
D
$aL^2 + bL^3$

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Work Done by Variable Force and Force-Displacement Graph

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DifficultJEE MAIN 2021

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When a rubber band is stretched by a distance $x$,it exerts a restoring force of magnitude $F = ax + bx^2$,where $a$ and $b$ are constants. The work done in stretching the unstretched rubber band by $L$ is

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The figure shows the relationship between force and position. What is the work done by the force for the displacement of the object from $x = 1 \; cm$ to $x = 5 \; cm$ in ergs?

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