JEE MAIN - Physics (2002 - No. 29)
A spring of force constant $$800$$ $$N/m$$ has an extension of $$5$$ $$cm.$$ The work done in extending it from $$5$$ $$cm$$ to $$15$$ $$cm$$ is
$$16J$$
$$8J$$
$$32J$$
$$24J$$
Explanation
When we extend the spring by $$dx$$ then the work done
$$dW = k\,x\,dx$$
Applying integration both sides we get,
$$\therefore$$ $$W = k\int\limits_{0.05}^{0.15} {x\,dx} $$
$$ = {{800} \over 2}\left[ {{{\left( {0.15} \right)}^2} - {{\left( {0.05} \right)}^2}} \right] $$
$$= 8\,J$$
$$dW = k\,x\,dx$$
Applying integration both sides we get,
$$\therefore$$ $$W = k\int\limits_{0.05}^{0.15} {x\,dx} $$
$$ = {{800} \over 2}\left[ {{{\left( {0.15} \right)}^2} - {{\left( {0.05} \right)}^2}} \right] $$
$$= 8\,J$$
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