A ball of mass 100 g is dropped from a height h =10 cm on a platform fixed at the top of vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance ( h /2). The spring constant is Nm -1 . (. Use .g=10 ms -2)

Question

A ball of mass 100 g is dropped from a height h =10 cm on a platform fixed at the top of vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance ( h /2). The spring constant is Nm -1 . (. Use .g=10 ms -2) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

By energy conservation PE = KE mg ( H +( H /2))=(1/2) kx 2( x =( H /2)) 0.100 × 10 × (3/2)(0.10) =(1/2) k (0.05 × 0.05) k =(3 × 0.10/0.05 × 0.05) =(3 × 1000/25)=120 N / m

Q. A ball of mass $100 g$ is dropped from a height $h = 10 c m$ on a platform fixed at the top of vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $\frac{h}{2}$ . The spring constant is _____ $N m^{- 1} . ($ Use $g = 10 m s^{- 2})$

By energy conservation
$PE = K E$
$m g (H + \frac{H}{2}) = \frac{1}{2} k x^{2} (x = \frac{H}{2})$
$0.100 \times 10 \times \frac{3}{2} (0.10)$
$= \frac{1}{2} k (0.05 \times 0.05)$
$k = \frac{3 \times 0.10}{0.05 \times 0.05}$
$= \frac{3 \times 1000}{25} = 120 N / m$

Q. A ball of mass 100g is dropped from a height h=10cm on a platform fixed at the top of vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance 2h​. The spring constant is _____Nm−1.( Use g=10ms−2)

By energy conservation PE=KE mg(H+2H​)=21​kx2(x=2H​) 0.100×10×23​(0.10) =21​k(0.05×0.05) k=0.05×0.053×0.10​ =253×1000​=120N/m

Q. A ball of mass $100 g$ is dropped from a height $h = 10 c m$ on a platform fixed at the top of vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $\frac{h}{2}$ . The spring constant is _____ $N m^{- 1} . ($ Use $g = 10 m s^{- 2})$

By energy conservation
$PE = K E$
$m g (H + \frac{H}{2}) = \frac{1}{2} k x^{2} (x = \frac{H}{2})$
$0.100 \times 10 \times \frac{3}{2} (0.10)$
$= \frac{1}{2} k (0.05 \times 0.05)$
$k = \frac{3 \times 0.10}{0.05 \times 0.05}$
$= \frac{3 \times 1000}{25} = 120 N / m$