In Figure, a block of mass $\mathbf{m}\mathbf{=}\mathbf{12}\mathbf{kg}$is released from rest on a frictionless incline of angle $\mathbf{30}\mathbf{°}$. Below the block is a spring that can be compressed 2.0 cmby a force of 270 N . The block momentarily stops when it compresses the spring by 5.5 cm. (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches the spring?

i) The initial velocity of the block is,u =0 m/s

ii) The inclination angle of the block is, $\theta =30°$

iii) The compression of the spring is, $x=2.0\mathrm{cm}=0.020\mathrm{m}$

iv) The force acting on the block is,F=270 N

v) The compression of the spring when block stops as d=5.5 c=0.055 m

Use the concept of the law of energy conservation and elastic potential energy of the spring. By using Hooke’s law, find the force constant. To find the height, use trigonometry.According to the law of energy conservation, energy can neither be created, nor be destroyed.

Formulae:

$$\begin{gathered} F=-kx \\ U\left(x\right)=\frac{1}{2}k{x}^2 \\ K=\frac{1}{2}m{v}^2 \end{gathered}$$

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.

The block is placed at point A as shown in the figure. It is at rest position. Hence, kinetic energy of the block is zero at that point. It comes in contact at point B. Due to the application of force, the spring gets compressed up to point C.

According to Hooke’s law,

$$\begin{gathered} F=kx \\ k=\frac{F}{x} \\ k=\frac{270N}{0.02m} \\ k=1.35\times {10}^4\mathrm{N}/\mathrm{m} \end{gathered}$$

The total distance covered by the block is I+d as shown in the figure. The initial height of the block is . By using trigonometry,

$$\begin{gathered} \sin \theta =\frac{h_A}{I+d} \\ I+d=\frac{h_A}{\sin \theta } \end{gathered}$$

(i)

The block moves from point A to point C, during this motion, the gravitational motion of the block is converted into elastic potential energy. Hence, according to the energy conservation law,

$mg{h}_A=\frac{1}{2}k{x}^2$

Here,x=d

$$\begin{gathered} mg{h}_A=\frac{1}{2}k{x}^2 \\ {h}_A=\frac{k{d}^2}{2mg} \\ {h}_{\mathrm{A}}=\frac{1.35\times {10}^4\mathrm{N}/\mathrm{m}\times {\left(0.055\mathrm{m}\right)}^2}{2\times 12\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2} \\ {\mathrm{h}}_{\mathrm{A}}=0.174\mathrm{m} \end{gathered}$$

The total distance travelled by the block, from the equation (1) is,

$$\begin{gathered} I=d=\frac{0.174}{\sin 30}m \\ I+d=0.347\mathrm{m} \\ \mathrm{I}+\mathrm{d}=0.35\mathrm{m} \end{gathered}$$

Hence, the total distance covered by the block before coming to stop is, $I+d=0.35m$

From the figure,

$$\begin{gathered} I+d=0.347m \\ I=0.347m-0.055\mathrm{m} \\ I=0.292\mathrm{m} \end{gathered}$$

The distance between point A to the point B is I=0.292 m . The vertical distance covered by the block from the point A to the point B is,

$$\begin{gathered} \Delta y=h_A-h_B \\ \Delta y=l\sin \theta \\ \Delta y=0.292m\times \sin 30 \\ \Delta y=0.146m \end{gathered}$$

According to the energy conservation law, the gravitational potential energy of the block is converted to kinetic energy and potential energy at point B,

$$\begin{gathered} 0+mg{h}_A=\frac{1}{2}m{v}_B^2+mg{h}_B \\ \frac{1}{2}m{v}_B^2=mg{h}_A-mg{h}_B \\ \frac{1}{2}m{v}_B^2=mg(h_A-h_B) \\ \frac{1}{2}{v}_B^2=g\Delta y \\ {v}_B^2=2g\Delta y \\ {v}_B=\sqrt{2g\Delta y} \\ {v}_B=\sqrt{2\times 9.8m/s^2\times 0.146m} \\ {\mathrm{v}}_{\mathrm{B}}=1.69\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{B}}=1.7\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the block just as it touches the spring is,${\mathrm{v}}_{\mathrm{B}}=1.7\mathrm{m}/\mathrm{s}$

Therefore, thetotal distance covered by the block and the velocity of the blockcan be found by using the concept of conservation of energy and elastic potential energy.