Question

Figure 8-36 shows an 8kgstone at rest on a spring. The spring is compressedby the stone.

(a) What is the spring constant?

(b) The stone is pushed down an additional 30 cm and released. What is the elastic potential energy of the compressed spring just before that release?

(c) What is the change in the gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height?

(d) What is that maximum height, measured from the release point?

Short answer

(a) Spring constant is $784N/m$

(b) Elastic potential energy of the compressed spring just before it released after pulling down is $62.7J$.

(c) Change in gravitational potential energy of system is $62.7J$.

(d) Maximum height reached by the stone after release is$0.8m$.

Step-by-step solution

Step 1: Given

1. Mass of stone, $m=8.0kg$
2. Compression of spring,$y=0.100m$

Determining the concept

The problem deals with the law of conservation of energy and Hooke’s law. According to the law of energy conservation, energy can neither be created, nor be destroyed.Hooke’s law states that the displacement or size of a deformation is directly proportional to the deforming force or load for relatively modest deformations of an object.Using the energy conservation law and Hooke’s law, find the spring constant, change in potential energy and maximum height reached by the stone after releasing it.

Formula:

Force is given by,

$f=ma$

Force by Hook’s law is given by,

$f=kx$

Elastic potential energy,

$U=\frac{1}{2}k{y}^2$

where, Uis potential energy, m is mass, v is velocity, a is an acceleration, x, y are displacements, k is spring constant, and F is force.

(a) Determining the spring constant

By Hooke’s law,

$F=kx$

And by Newton’s law,

$f=ma$

So,

$$\begin{gathered} kx=mg \\ k\left(0.10\right)=8\left(9.8\right) \\ k=784\text{N/m} \end{gathered}$$

Hence, spring constant is $784N/m$.

(b) Determining the elastic potential energy of compressed spring just before it is released

If the spring is compressed for 30.0 cm again, the total compression is,

So, as per the elastic potential energy formula,

$$\begin{gathered} U=\frac{1}{2}k{y}^2 \\ U=\frac{1}{2}\left(784\right){\left(0.400\right)}^2 \\ U=\frac{1}{2}\left(784\right){\left(0.400\right)}^2 \\ U=62.7J \end{gathered}$$

Hence, elastic potential energy of the compressed spring just before it released after pulling down is $62.7J$.

(c) Determining the change in gravitational potential energy of the system

As, thestone is released from thecompressed spring, it has potential energy, during movement, it has kinetic energy, and at thehighest point, it has only potential energy as the highest point doesn’t have velocity. So, energy will remain constant. That is,$62.7J$

Hence, change in gravitational potential energy of system is $62.7J$.

(d) Determining the maximum height reached by the stone after the release

At maximum height, the stone has only potential energy. So,

$$\begin{gathered} mgh=62.72J \\ \left(8\right)\left(9.8\right)h=62.72J \\ h=0.8m \end{gathered}$$

Hence, maximum height reached by the stone after release is $0.8m$.