Question

*The graph below shows the force exerted by a spring as a function of the length of the spring. A block on a frictionless table is pushed against the spring that is fastened to a wall. The spring's rest length is 30 cm. The block is pressed against the spring until its length becomes 10 cm. The block is then released.

Answer 1

Given data:

The spring's rest length is x=30 cm.

The compressed length of spring is x'=10 cm.

The displacement in the spring will be,

`\begin{gathered} \Delta x=x-x^(\prime) \\ \Delta x=30\text{ cm}-10\text{ cm} \\ \Delta x=20\text{ cm}*\frac{1\text{ m}}{100\text{ cm}} \\ \Delta x=0.2\text{ m} \end{gathered}`

The slope of the given graph will be equal to the spring stiffness that can be calculated as,

`\begin{gathered} k=(60-0)/(0.3-0.2) \\ k=600\text{ N/m} \end{gathered}`

The energy observed by the spring will be equal to the kinetic energy of the block. We can equate both to calculated the kinetic energy of the block,

`\begin{gathered} \text{Kinetic energy=Spring energy} \\ KE=(1)/(2)k\Delta x^2 \\ KE=(1)/(2)(600)(0.2)^2 \\ KE=12\text{ J} \\ \\ \end{gathered}`

Thus, the kinetic energy of the block is 12 J.