What is the spring constant (in n/m) of a spring that stores 100 J of elastic potential energy when it is stretched 5 m?

8N/m

Explanation:

The elastic potential energy stored can be calculated using the equation:

elastic potential energy = 0.5 × spring constant × (extension)2

assuming it is not stretched beyond the limit of proportionality.

Then calculate using the values in the question:

100J = 0.5 × spring constant × 5²

200J = spring constant × 25

spring constant = 200J ÷ 25m²

spring constant = 8J/m²

spring constant = 8N/m

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j