# StudySquare

This page covers the following topics:

1. Gravitational potential energy

2. Elastic potential energy

3. Elastic strain energy

4. Kinetic energy

5. Deriving energy equations

Gravitational potential energy refers to the energy possessed by an object due to its position above the ground. It can be calculated by multiplying the mass of an object, gravitational field strength and height the object has been lifted to. Gravitational field strength on the surface of the Earth is commonly approximated to be 9.81 m/s².

Elastic potential energy is the energy stored in elastic objects when they are stretched or compressed. This occurs when a force is applied to an object, which causes it to change shape. When the object returns to its original shape, the stored elastic potential energy is released.

All the work done in stretching or compressing an elastic object is stored as elastic strain energy, which is equivalent to the elastic potential energy. The elastic strain energy in J can be calculated using E = 1/2 × F × x, where F is the force that is causing the stretching or compressing of the object in N, and x is the extension in m. This energy is equivalent to the area under a force-extension graph.

Kinetic energy is the energy an object has due to its motion and thus its speed. It can be calculated using E = 1/2 × mass × velocity² and is given in joules, J. A moving vehicle can be stopped only when work is done on it which is equivalent to the kinetic energy the vehicle possesses.

The equation for gravitational potential energy is derived using the first part of the diagram. Consider an object in equilibrium of weight mg. Since the object is in equilibrium, the resultant force on it must be 0, thus there is an upwards force of mg acting on the object. When the object is moved through a distance of Δh, work is done on it. Using the equation for work done, WD = mgΔh. Since the object is changing position, this work done will be transferred into a gain of gravitational potential energy and therefore ΔGPE = mgΔh, as required.

The equation for kinetic energy is derived using the second part of the diagram. Consider a moving object. The work done by the object will be equal to its kinetic energy, therefore WD = Fs = ΔKE, where F is the force exerted by the object. Using F = ma from Newton's second law, we get ΔKE = mas. Using an equation from kinematics, v² = u² + 2as, and rearranging gives as = (v² − u²)/2. Substituting this into the expression for kinetic energy gives ΔKE = m(v² − u²)/2 = mΔv²/2, as required.

# 1

Kate drops a 275 g volleyball towards the ground. Given that when the ball is 0.5 m above the ground the change in gravitational potential energy is 2.7 J, calculate the height above the ground at which the ball is dropped from. Use g = 9.81 N/kg.

E = mgh

2.7 J = 0.275 kg × 9.81 N/kg × change in height

change in height = 2.7 J/(0.275 kg × 9.81 N/kg) = 1.00 m (3 s. f.)

height above the ground initially = 0.5 m + 1.00 m = 1.5 m

1.5 m

# 2

Calculate the gravitational potential energy stored in a tennis ball of mass 60 g that is held 1.4 m above the ground. Use g = 9.81 N/kg.

E = mgh

E = 0.06 kg × 9.81 N/kg × 1.4 m = 0.82 J

0.82 J

# 3

A spring is stretched by 10 cm using a force of 10 N. Calculate the elastic strain energy stored in the spring.

E = (1/2)Fx

E = (1/2) × 10 N × 0.1 m = 0.5 J

0.5 J

# 4

What is the necessary property of an object in the derivation of the gravitational potential energy?

The object must be in equilibrium, so that the fact that the resultant force acting on it being zero can be used to find the upwards force acting on it.

The object must be in equilibrium.

# 5

A biker is driving his bike of mass 10 kg at a speed of 7.2 m/s. Given that the kinetic energy of the bike and biker is 1763 J, calculate the mass of the biker.

E = 0.5mv²

Let m be the biker's mass.

1763 J = 0.5 × (m + 10 kg) × (7.2 m/s)²

m = (2 × 1763) ÷ 7.2² − 10 = 58 kg

58 kg

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