Aspring has a force constant of 500.0 n/m. show that the potential energy stored in the spring is as follows: a. 0.400 j when the spring is stretched 4.00 cm from equilibrium b. 0.225 j when the spring is compressed 3.00 cm from equilibrium c. zero when the spring is unstretched

(a) Hence, the potential energy = 0.400 J.

(b) Hence the potential Energy = 0.225 J.

(c) Hence the potential energy = 0 J

Explanation:

Potential Energy; This is the energy of a body, by virtue of its position in the gravitational field. The unit of potential energy is Joules (J)

the potential energy stored in a spring is

Ep = 1/2ke² Equation 1

Where Ep = potential Energy, k = force constant,  e = extension.

a.

When spring is stretched 4.00 cm,  e = 4.00 cm = 4/100 = 0.04 m, and k = 500 N/m.

Substituting into equation 1

Ep = 1/2(500)(0.04)²

Ep = 0.4 J.

Hence, the potential energy = 0.400 J.

(b)

When the spring is compressed 3.00 cm, e = 3.00 cm = 3/100 = 0.03 m, and k = 500 N/m.

Substitute into equation 1,

Ep = 1/2(500)(0.03)²

Ep = 250(0.0009)

Ep = 0.225 J.

Hence the potential Energy = 0.225 J.

(c) When the spring is unstretched, e = 0 cm = 0 m and k = 500 N/m.

Substituting into equation 1

Ep = 1/2(500)(0)²

Ep = 250(0)

Ep = 0.

Hence the potential energy = 0 J

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s