Question

A 1.0 kg block is attached to an unstretched
spring of spring constant 50 N/m and released
from rest from the position shown in Figure 1.
The block oscillates for a while and eventually
stops moving 0.20 m below its starting point, as
shown in Figure 2. What is the change in
potential energy of the block-spring-Earth
system between Figure 1 and Figure 2?

Answer 1

The change in the potential energy of the block-spring-Earth system is 1 J.

What is elastic potential energy?

The stored energy in elastic due to the compression or stretch in the elastic is known as elastic potential energy.

From the formula of elastic potential energy:

$U = \dfrac 12 kx ^2$

Where,

$k$- spring constant = 50 N/m

$x$ - displacement = 0.2 m

Since the spring was at rest initially,

So, $Ug = 0$

The potential energy after release,

$Us = \dfrac 12 50(0.2)^2\\Us = \dfrac 12 50\times 0.04\\Us = 1$

Now the potential difference,

$\Delta U = Us - Ug$

$\Delta U =1 - 0\\\\\Delta U = 1 \rm \ J$

Therefore, the change in  potential energy of the block-spring-Earth  system is 1 J.

To know more about potential difference,

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