A ball of mass $$100$$ g is dropped from a height $$h = 10$$ cm on a platform fixed at the top of a 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$$^{-2}$$)

Given,

• Mass, m=100 g ,=0.1 kg ,m = 100
• Height, h=10 cm, h =0.1 
• $$g = 10$$ m s$$^{-2}$$
• Compression of spring =$$\ \frac{\ h}{2}$$=0.05
  
  The ball is dropped from a height  above the platform. After reaching the platform, it continues to move downward and compresses the spring by an additional distance$$\ \frac{\ h}{2}$$

Thus, the total downward displacement of the ball from its initial position to the point of maximum compression is $$h\ +\ \ \frac{\ h}{2}\ =\ \ \frac{\ 3h}{2}$$

During this motion, the ball loses gravitational potential energy corresponding to this total displacement.

P.E = $$m\times\ g\times\ h_{total}$$

P.E = $$m\times\ g\times\ \ \frac{\ 3h}{2}$$

At the point of maximum compression, the velocity of the ball becomes zero, and all the lost gravitational potential energy is stored as elastic potential energy in the spring.

$$U=\ \frac{\ 1}{2}\times\ k\times\ x^2$$

$$U=\ \frac{\ 1}{2}\times\ k\times\ \left(\frac{\ h}{2}\right)^2$$

Applying conservation of energy, the loss in gravitational potential energy is equal to the gain in spring potential energy.

P.E = U
$$m\times\ g\times\ \ \frac{\ 3h}{2} = \ \frac{\ 1}{2}\times\ k\times\ \left(\frac{\ h}{2}\right)^2$$ 

Now substitute the given values of mass, gravity, and height into the equation.

$$0.1\times\ 10\times\ 0.15\ =\ \ \frac{\ 1}{2}\times\ k\times\ \left(\ \frac{\ 0.1}{2}\right)^2$$

On simplifying the expression, we obtain the value of the spring constant.

$$\ 0.15\ =\ k\frac{\ 0.01}{8}$$

$$k\ =\ \ \frac{\ 1.2}{0.01}$$

$$k\ =\ \ \ 120 N m ^{-1}$$

Hence, the spring constant is  120 N m$$^{-1}$$