Question

Find the mechanical energy of a block–spring system having a spring constant 1.3 N/ cmofand oscillation amplitude of 2.4cm.

Short answer

The mechanical energy of the block-spring system is$3.74\times {10}^{-2}\mathrm{J}$

Step-by-step solution

The given data

1. The amplitude of motion,${\mathrm{x}}_{\mathrm{m}}=2.4\mathrm{cm}\mathrm{or}0.024\mathrm{m}$
2. The spring constant of the system,$\mathrm{k}=1.3\mathrm{N}/\mathrm{cm}\mathrm{or}130\mathrm{N}/\mathrm{m}$.

Understanding the concept of mechanical energy

We can find the mechanical energy of the block-spring system using the law of conservation of energy.

Formula:

The elastic P.E energy of the system $PE=\frac{1}{2}k{x}^2....\left(1\right)$

The law of conservation of energy gives E= constant (2)

Calculation of mechanical energy of the spring-block system

From equation (ii), we can say that the mechanical energy of the system = the maximum elastic P.E energy of the system. Hence, we get the total energy of the system as:

$$\begin{gathered} E=\frac{1}{2}k{x}_m^2 \\ =\frac{1}{2}130{(0.024)}^2=3.74\times {10}^{-2}J \end{gathered}$$

Therefore, the mechanical energy of the block-spring system is$3.74\times {10}^{-2}\mathrm{J}$.