A1.0 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?

Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = 1 Nm.

Explanation:

Here, spring constant, k  = 50 N/m.

given block comes down eventually 0.2 m below.

here, g = 10 m/s.

let block be at a height h above the ground in figure 1.

⇒In figure 2, potential energy of the block-spring-Earth

system = m×g×(h - 0.2) + 1/2× k × x². where, x = change in spring length.

⇒ Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = (m×g×(h - 0.2)) - (1/2× k × x²)

              =  (1×10×0.2) - (1/2×50×0.2×0.2) = 1 Nm.

real, smaller and upside-down.   so the answer should be b.