91Ó°ÊÓ

i) The mass of the block is $m=2.00\mathrm{kg}$

ii) The initial velocity of the block is $u=0\mathrm{m}/\mathrm{s}$

iii) The inclination angle of the block is $\mathrm{θ}=30.0°$

iv) The compression of the spring is$x=20.0\mathrm{cm}=0.20\mathrm{m}$

v) The spring constant of the spring is,$k=19.6\frac{\mathrm{N}}{\mathrm{cm}}=1960\mathrm{N}/\mathrm{m}$

Use the concept of the energy conservation law and elastic potential energy of the spring. To find the distance alongtheincline, use trigonometry.According to the law of energy conservation, energy can neither be created, nor be destroyed.

Formulae:

F =-kx

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

$K=\frac{1}{2}m{v}^2$

where, K is kinetic energy, U(x) is potential energy, m is mass, v is velocity, x is displacement, k is spring constant and F is force.

Initially, the spring is compressed by the block. The elastic potential energy of the block is,

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

$U\left(x\right)=\frac{1}{2}×1960\mathrm{N}/\mathrm{m}×(0.200\mathrm{m}{)}^2$

$U\left(x\right)=39.2\mathrm{J}$

Hence, the elastic potential energy of the compressed spring is $U\left(x\right)=39.2\mathrm{J}$.

The spring is compressed by the block and then released. After that, the block separates from the spring and goes at the highest point and stops. During compression, the spring has elastic potential energy. According to the energy conservation law, elastic potential energy is converted to gravitational potential energy.

$$\begin{gathered} \mathrm{Δ}U=U\left(x\right) \\ \mathrm{Δ}U=39.2J \end{gathered}$$

Hence, the change in gravitational potential energy of the system is $\mathrm{Δ}U=39.2J$.

The block is starting fromtherest and it stops at the highest point.Hence, its initial and final kinetic energy is zero. According to the mechanical energy conservation law,

$K_0+U_0=K_f+U_f$

$0+\frac{1}{2}k{x}^2=0+mgh$

$h=\frac{{\displaystyle \frac{1}{2}}k{x}^2}{m}$

$$\begin{gathered} h=\frac{{\displaystyle \frac{1}{2}}×1960\mathrm{N}/\mathrm{m}×(0.200\mathrm{m}{)}^2}{2.00\mathrm{kg}} \\ \mathrm{h}=2.00\mathrm{m} \end{gathered}$$

The distance covered by the block along the incline plane by using trigonometry is,

$\sin \mathrm{θ}=\frac{h}{d}$

$d=\frac{h}{\sin \mathrm{θ}}$

$d=\frac{2.00\mathrm{m}}{\sin 3.0}$

$d=4.00\mathrm{m}$

Hence, the distance covered by the block along the inclined plane is d =4.00 m .

Therefore, the elastic potential energy, the change in gravitational potential energy and the total distance covered by the block can be found using the concept of conservation of energy and elastic potential energy.