Question

A spring is hung from the ceiling. A 0.350-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.130 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring?

Answer 1

Answer:

K = 26.92 N/m  

Explanation:

Given that

m = 0.35 kg

x= 0.13 m

Lest take spring constant = K

When spring stretch by 0.13 m,then the spring force

F= K x

F= 0.13 K

For balancing the spring force ,the gravity force = m g

Therefore

m g= K x      

Now by putting the values in the above expression  

0.35 x 10 = 0.13 K                       ( take g =10 m/s²)

K = 26.92 N/m  

Therefore the spring constant = 29.92 N/m