A ball of mass 100 , g is dropped from a heig | vedclass

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(D) Let the mass of the ball be $m = 100 \, g = 0.1 \, kg$. The height from which it is dropped is $h = 10 \, cm = 0.1 \, m$. The compression in the s

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(D) Let the mass of the ball be $m = 100 \, g = 0.1 \, kg$. The height from which it is dropped is $h = 10 \, cm = 0.1 \, m$. The compression in the spring is $x = \frac{h}{2} = 5 \, cm = 0.05 \, m$.By the principle of conservation of mechanical energy,the loss in gravitational potential energy of the ball equals the gain in elastic potential energy of the spring.The total vertical displacement of the ball is $h + x = h + \frac{h}{2} = \frac{3h}{2}$.Therefore,$mg \left( h + \frac{h}{2} ight) = \frac{1}{2} kx^2$.Substituting the values: $0.1 	imes 10 	imes \left( 0.1 + 0.05 ight) = \frac{1}{2} 	imes k 	imes (0.05)^2$.$1 	imes 0.15 = \frac{1}{2} 	imes k 	imes 0.0025$.$0.15 = k 	imes 0.00125$.$k = \frac{0.15}{0.00125} = \frac{150000}{1250} = 120 \, N \, m^{-1}$.

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