JEE MAIN - Physics (2003 - No. 63)
A spring of spring constant $$5 \times {10^3}\,N/m$$ is stretched initially by $$5$$ $$cm$$ from the unstretched position. Then the work required to stretch it further by another $$5$$ $$cm$$ is
$$12.50$$ $$N$$-$$m$$
$$18.75$$ $$N$$-$$m$$
$$25.00$$ $$N$$-$$m$$
$$625$$ $$N$$-$$m$$
ব্যাখ্যা
Given $$k = 5 \times {10^3}N/m$$
Work done when a spring stretched from x1 cm to x2 cm,
$$W = {1 \over 2}k\left( {x_2^2 - x_1^2} \right) $$
$$= {1 \over 2} \times 5 \times {10^3}\left[ {{{\left( {0.1} \right)}^2} - {{\left( {0.05} \right)}^2}} \right]$$
$$ = {{5000} \over 2} \times 0.15 \times 0.05 = 18.75\,\,Nm$$
Work done when a spring stretched from x1 cm to x2 cm,
$$W = {1 \over 2}k\left( {x_2^2 - x_1^2} \right) $$
$$= {1 \over 2} \times 5 \times {10^3}\left[ {{{\left( {0.1} \right)}^2} - {{\left( {0.05} \right)}^2}} \right]$$
$$ = {{5000} \over 2} \times 0.15 \times 0.05 = 18.75\,\,Nm$$