The potential energy stored in the compressed spring of a dart gun, with a spring constant of 31.50 N/m, is 1.240 J. Find by how much is the spring is compressed.

0.281  m

Explanation:

Applying,

W = 0.5ke² Equation 1

Where W = Workdone by the stretched spring, k = spring constant, e = extension/ compression of the spring

make e the subject of the equation

e = √(2W/k) Equation 2

From the question,

Given: W = 1.240J, k = 31.50 N/m

Substitute these values into equation 2

e = √(2×1.240/31.50)

e = √(0.07873)

e = 0.281  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