High School

An object of mass 3 kg is hung from a spring with a spring constant of [tex]114 \, \text{N/m}[/tex]. Calculate the elastic potential energy stored by the spring when the object is at rest. Assume the gravitational field strength is [tex]9.8 \, \text{N/kg}[/tex].

Answer :

Answer:

3.79 J

Explanation:

The elastic potential energy (EE) stored in the spring is equal to half of the spring constant (k) times the square of the displacement (Δx). To find how far the spring is displaced, we can use Hooke's law. This law says that the force on the spring (F) is equal to the product of the spring constant (k) and the displacement (Δx).

Hooke's Law

First, let's use Hooke's law to find how far the spring is displaced. The force on the spring is equal to the weight of the object hanging from it (mg).

[tex]\Large \text {$ F=k\Delta x $}\\\\\Large \text {$ mg=k\Delta x $}\\\\\Large \text {$ (3\ kg)(9.8\ N/kg)=(114\ N/m)\Delta x $}\\\\\Large \text {$ \Delta x=0.258\ m $}[/tex]

Elastic Energy

Now we can find the amount of elastic potential energy.

[tex]\Large \text {$ EE=\frac{1}{2}k{\Delta x}^2 $}\\\\\Large \text {$ EE=\frac{1}{2}(114\ N/m)(0.258\ m)^2 $}\\\\\Large \text {$ EE=3.79\ J $}[/tex]