# Potential energy – problems and solutions

**Gravitational potential energy**< /p>

1. Energy 4900 Joule used to raise an object with mass of 50 kg to a height of h. What is the height of h? Acceleration due to gravity

^{-2}.

__Known :__

Change of potential energy (ΔPE) = 4900 Joule

Mass of object (m) = 50 kg

Acceleration due to gravity (g) = 9.8 m/s^{2}

__Wanted:__ height (Δh)

__Solution :__

ΔPE = m g Δh

4900 = (50)(9.8) Δh

4900 = 490 Δh

Δh = 10 meters

2. The graph below shows the relation between force (F) and x (the change in length) of a spring. If the change in length of a spring is 8 cm, what is the spring potential energy?

__Known :__

Force (F) = 2 Newton

The change in length 1 (x) = 1 cm = 1/100 m = 0.01 m

The change in length 2 = 8 cm = 8/100 m = 0.08 m

__Wanted :__ The spring potential energy

__Solution :__

Spring’s constant :

The spring potential energy :

∆PE = 1/2 k x^{2}

∆PE = 1/2 (200 N/m)(0.08 m)^{2}

∆PE = (100 N/m)(0.0064 m^{2})

∆PE = 0.64 Nm

3. Based on table below, F = weight of object, ∆L = the change in length of spring. What is the work done on the spring so the change in length of spring is 10 cm.

__Known :__

The change in length of spring (∆L) = 10 cm = 0.1 m

__Wanted :__ Work done on the spring

__Solution :__

Spring’s constant :

k = F / ∆x = 20 N / 0.04 m = 500 N/m

k = F / ∆x = 30 N / 0.06 m = 500 N/m

k = F / ∆x = 40 N / 0.08 m = 500 N/m

Spring’s constant is 500 N/m

The work done on the spring so the change in length of spring is 10 cm :

W = 1/2 k ∆L^{2} = 1/2 (500 N/m)(0.1 m)^{2} = (250 N/m)(0.01 m^{2}) = 2.5 N m = 2.5 Joule.

4. Graph below shows relation between force (F) and the change in length (x). What is the spring potential energy based on graph below.

__Known :__

F = 40 Newton

Δx = 0.08 m

__Wanted :__ spring potential energy

__Solution :__

Spring’s constant :

k = F / Δx

The spring potential energy :

PE = ½ k Δx^{2 }= ½ (F/Δx) Δx^{2 }= ½ F Δx

PE = ½ (40)(0.08) = (20)(0.08) = 1.6 Joule