JEE MAIN - Physics (2014 (Offline) - No. 27)
When a rubber-band is stretched by a distance $$x$$, it exerts restoring force of magnitude $$F = ax + b{x^2}$$ where $$a$$ and $$b$$ are constants. The work done in stretching the unstretched rubber-band by $$L$$ is :
$$a{L^2} + b{L^3}$$
$${1 \over 2}\left( {a{L^2} + b{L^3}} \right)$$
$${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$$
$${1 \over 2}\left( {{{a{L^2}} \over 2} + {{b{L^3}} \over 3}} \right)$$
설명
Given Restoring force, F = ax + bx2
Work done in stretching the rubber-band by a distance $$dx$$ is
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dW = F\,dx = \left( {ax + b{x^2}} \right)dx$$
Intergrating both sides,
$$W = \int\limits_0^L {axdx + \int\limits_0^L {b{x^2}dx}}$$
= $$\left[ {a{{{x^2}} \over 2} + b{{{x^3}} \over 3}} \right]_0^L$$
= $${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$$
Work done in stretching the rubber-band by a distance $$dx$$ is
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dW = F\,dx = \left( {ax + b{x^2}} \right)dx$$
Intergrating both sides,
$$W = \int\limits_0^L {axdx + \int\limits_0^L {b{x^2}dx}}$$
= $$\left[ {a{{{x^2}} \over 2} + b{{{x^3}} \over 3}} \right]_0^L$$
= $${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$$