Question

A block whose weight is 48 newtons slides from rest down a frictionless plane that is inclined 30° to the horizontal.It then moves across a level frictionless surface and collides with a perfect spring. The block compresses the springfrom its equilibrium position before coming momentarily to rest. The spring constant is 3270 newtons per meter andthe length of the plane is 24 meters, as shown. Neglect any energy losses due to friction during the collision.(a) Find the maximum distance that the spring was compressed from equilibrium.(b) Find the speed of the block just before it collides with the spring.

Answer 1

Given data

The weight of the block is W = 48 N

The angle of inclination is theta = 30o

The spring constant is k = 3270 N/m

The length of the plan is s = 24 m

(a)

The expression for the energy at the top of the inclination is given as:

`PE=W* s*\sin \theta`

The expression for the kinetic energy stored in the block due to the block moving down the plane is given as:

`\begin{gathered} KE=(1)/(2)mv^2 \\ KE=(1)/(2)(W)/(g)v^2 \end{gathered}`

From the energy conservation principle, the expression is written as follows:

`\begin{gathered} KE=PE \\ (1)/(2)(w)/(g)v^2=W* s*\sin 30^o \\ v=\sqrt[]{2gs\sin \theta} \end{gathered}`

Substitute the value in the above equation.

`\begin{gathered} v=\sqrt[]{2*9.8m/s^2*24\text{ m}*\sin 30\circ} \\ v=15.33\text{ m/s} \end{gathered}`

The expression for the maximum distance that the spring was compressed from equilibrium is given as:

`\begin{gathered} U_s=KE \\ (1)/(2)kx^2=(1)/(2)(W)/(g)* v^2 \\ x=\sqrt{(Wv^2)/(gk)} \end{gathered}`

Substitute the value in the above equation.

`\begin{gathered} x=\sqrt[]{\frac{48\text{ N}*(15.33)^2}{9.8m/s^2*3270\text{ N/m}}} \\ x=0.593\text{ m} \end{gathered}`

Thus, the maximum distance that was compressed from equilibrium is 0.593 m.

(b)

Thus, the speed of the block just before it collides with the spring is 15.33 m/s.