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

  • A
    $\frac{1}{2}(aL^2 + bL^3)$
  • B
    $\frac{aL^2}{2} + \frac{bL^3}{3}$
  • C
    $\frac{aL^2}{3} + \frac{bL^3}{5}$
  • D
    $\frac{aL^2}{7} + \frac{bL^3}{9}$

Explore More

Similar Questions

The relationship between the force $F$ and position $x$ of a body is as shown in the figure. The work done in displacing the body from $x = 1 \text{ m}$ to $x = 5 \text{ m}$ will be ........ $J$.

The figure shows the variation of a force $F$ acting on a particle along the $x$-axis. If the particle begins at rest at $x=0$,what is the particle's coordinate when it again has zero speed?

$A$ block of mass $10 \, kg$,moving in $x$ direction with a constant speed of $10 \, m/s$,is subjected to a retarding force $F = 0.1 \, x \, N$ during its travel from $x = 20 \, m$ to $30 \, m$. Its final $KE$ will be ............... $J$.

If a position-dependent force $F(x) = (3x^2 - 2x + 7) \text{ N}$ acting on a body of mass $2 \text{ kg}$ displaces it from $x = 0 \text{ m}$ to $x = 5 \text{ m}$,then the work done by the force is: (in $J$)

The force $F$ acting on a particle moving in a straight line is shown below. What is the work done by the force on the particle in the $1^{\text{st}}$ meter of the trajectory (in $\,J$)?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo