Question

A slab of mass m is released from a height $h_0$ from the top of a spring S' of force constant k. The maximum compression x of the spring is given by the equation

• A. $mg{h}_0=\left(\frac{1}{2}\right)k{x}^2$
• B. $mg(h_0-x)=\left(\frac{1}{2}\right)k{x}^2$
• C. $mg{h}_0=\frac{1}{2}k(h_0+x{)}^2$
• D. $mg(h_0+x)=\frac{1}{2}k{x}^2$

Answer 1

Answer: (D)

$mg(h_0+x)=\frac{1}{2}k{x}^2$

Potential energy of a spring is the energy associated with the state of compression or expansion of an elastic spring.
A definite amount of work has to be done against the restoring force, both in the compressing or in stretching the spring. This work done is stored in the spring in the form of elastic potential energy of the spring. Thus, potential energy of a spring is energy associated with the state of compression or expansion of an elastic spring.
So, loss in PE of the mass == potential energy stored in the compressed spring.
If x be the maximum compression produced in the spring, then loss in PE of the mass (which moves downwards through a distance$(ho+x)$$=mg(ho+x)$ (h0+x)
and potential energy stored in the compressed spring$=\frac{1}{2}k{x}^2$=12kx2
Hence, according to law of conservation of energy $mg(ho+x)=\frac{1}{2}k{x}^2$

mg(h0+x)=12kx2$mg(ho+x)=12kx^2$