Question

A stone with mass of 0.1 kg is catapulted as shown in the figure. The total force $F_x$ (in N) exerted by the rubber band as a function of distance x (in m) is given by $F_x=300x^2$. If the stone is displaced by 0.1 m from the un-stretched position the rubber band is

• A. 0.01 J
• B. 0.1 J
• C. 1 J
• D. 10 J

Answer 1

The correct option is B 0.1 J

Rate of change of energy stored $=Fdx$
$dE=300x^{2}dx$
$\int dE={\int }_0^{0.1}300x^{2}dx$
$E=300\times \frac{x^3}{3}{|}_0^{0.1}=0.1J$