Question

A ball of mass m is dropped from a height h on a massless platform fixed at the top of a vertical spring as shown below. The platform is depressed by a distance x. What will be the value of the spring constant?
answerrr it fastt....

Answer 1

Given

m = Mass of block

h = Displacement of platform

x = Displacement of spring

To find

Spring constant (k)

Solution

g = Acceleration due to gravity

The potential energy of the object and potential energy of the spring balance each other. In other words the energy of the system is conserved.

$mg(h+x)=\dfrac{1}{2}kx^2$

$\Rightarrow k=\dfrac{2mg(h+x)}{x^2}$

Therefore the spring constant is $\boldsymbol{\dfrac{2mg(h+x)}{x^2}}$.

Answer 2

Answer:

The particle is dropped from height h and the spring be compressed by x.

Loss in PE of the particle = gain in elastic PE of the spring

mg(h+x)= 1/2 kx^2 ⇒ k = 2mg(h+x)/x^2