10.2 - kinetic + potential energy

what is kinetic energy?

• the energy of an object due to its motion

• Ek = ½ m v²

how does speed vary with kinetic energy?

the faster an object moves, the more kinetic energy it has

what is the equation for kinetic energy?

Ek = ½ m v²

• Ek = kinetic energy

• m = object mass

• v² = speed squared

Ek = ½ m v²

kinetic energy

why is the equation for kinetic energy Ek = ½ m v²?

derivation here

when does the equation for kinetic apply?

at speeds below the speed of light

when doesn’t the equation for kinetic energy apply?

at speeds close to / at / higher than the speed of light

why doesn’t the kinetic energy equation apply to speeds near the speed of light?

because einstein’s theory of relativity (E = m c²) applies instead

how can energy be worked out at speeds near the speed of light?

using einstein’s theory of relativity, E = m c²

what is (gravitational) potential energy?

• the energy of an object due to its position

• m g Δ h

what is the equation for gravitation potential energy?

m g Δ h

• m g = object weight

• Δ h = change in height, height raised

why is the equation for GPE m g Δ h?

• work done = force x distance moved in the direction of the force

• GPE is measured as vertical height

• when being raised through a vertical height at a steady speed, the force needed is equal and opposite (balanced) to its weight, m g

• therefore work done = GPE = m g Δ h

how does the work done effect the GPE?

the work done increases its GPE

how can you increase GPE?

by doing work (to raise the object higher)

what is the equation for change of GPE?

Δ E_P = m g Δ h

• Δ E_P = change in potential energy

• m g = object mass

• Δ h = vertical distance raised

Δ E_P = m g Δ h

change of GPE

what is the energy transfer for an object in free fall?

½ m v² = m g Δ h

why is the energy transfer for an object free falling from a vertical height ½ m v² = m g Δ h?

• object’s energy when raised vertically is GPE

• object gains speed as it falls

• GPE decreases as it gets closer to the surface, E_k increases at it gains speed

• energy is conserved, therefore E_k = loss of GPE

• therefore ½ m v² = m g Δ h

when can ½ m v² = m g Δ h be applied?

when Δ h « earth’s radius

when can’t ½ m v² = m g Δ h be applied?

when Δ h < / > (of similar or greater magnitude) earth’s radius

why can ½ m v² = m g Δ h be applied when Δ h « earth’s radius?

because the object will still accelerate due to earth’s gravitational field strength (g) at this height, so the equation can be applied

why can’t v² = m g Δ h be applied when Δ h < / > (of similar or greater magnitude) earth’s radius?

because the value of g is not the same over the height h, and the force of gravity decreases with increased distance from the earth

what happens to the force of gravity with increased distance from the earth?

it decreases

when is a pendulum at GPE_max?

at max height, h₀

when is a pendulum at GPE_min?

at equilibrium position

when is a pendulum at Ek_max?

at equilibrium position

when is a pendulum at Ek_min?

at max height, h₀

does the GPE of a pendulum bob ever reach zero?

does the E_k of a pendulum bob ever reach zero?

yes

when does the E_k of a pendulum bob reach zero?

at max height, h₀

why does a pendulum bob reach zero E_k at max height h₀?

because the pendulum bob stops momentarily (has zero speed) when changing direction

what is the max height position of a pendulum?

h₀

what is the thread like when the pendulum bob is released?

taut

what does h₀ denote for a pendulum?

height at max height

what does h denote for a pendulum?

the height at any point above equilibrium and below max height (equilibrium height < h < max height)

what is the energy transfer in a pendulum?

½ m v² = m g (h₀ - h)

• 1 / 2 m v² = pendulum E_k

• m g (h₀ - h) = pendulum GPE

• h₀ = height at max height

• h = equilibrium height < h < max height

here

why is the energy transfer in a pendulum ½ m v² = m g (h₀ - h) ?

• the energy of a pendulum bob at max height is GPE. it has no E_k here because its either released from rest, or because the pendulum stops momentarily when turning

• pendulum bob speeds up when displaced from equilibrium. it passes equilibrium position at Ek_max and GPE_min

• it slows down to reach max height at the other side of the equilibrium position

• energy is conserved, so E_k = loss of GPE from max height

• h = Δ h = max height - height at any point (above equilibrium, below h₀) = h₀ - h

• therefore ½ m v² = m g (h₀ - h)

½ m v² = m g (h₀ - h)

energy transfer for a pendulum bob

what is the energy at different points of a pendulum bob’s motion?

here

what is the energy of a pendulum bob at max height?

• GPE_max

• Ek_min

what is the energy of a pendulum bob at rest?

• if displaced from equilibrium at h₀, GPE

• if displaced from equilibrium at h, GPE

• if at equilibrium but static, very little GPE

what is the energy of a pendulum bob at equilibrium?

• GPE_min

• Ek_max

why is the GPE_max of a pendulum bob at h₀?

• this is the highest vertical distance of the pendulum bob, therefore has its maximum value for GPE here

• none of the energy is stored as E_k at h₀ because the pendulum bob stops momentarily when turning

why is the GPE_min of a pendulum bob at equilibrium?

• this the lowest vertical distance of the pendulum bob, therefore has its minimum value for GPE here

• all the GPE has been converted to E_k at this point

why is the Ek_min of a pendulum bob at h₀?

• the pendulum bob is static as it stops momentarily whilst turning

• all energy is stored as GPE at h₀ cuz there is no motion and the highest vertical distance reached by the pendulum bob is at this point

why is the Ek_max of a pendulum bob at equilibrium?

the pendulum bob speeds up after being released at h₀, therefore passing the equilibrium with max speed as all the GPE has been transferred to E_k

what happens to the E_k of a pendulum bob after passing equilibrium?

it begins to slow down as it reaches h₀ on the other side of the equilibrium

what is the energy of a pendulum bob at h?

½ m v² = m g (h₀ - h)

what is the energy transfer in a fairground ride?

½ m v² = m g h

• ½ m v² = fairground ride E_k

• m g h = fairground ride GPE

½ m v² = m g h

• energy transfer in a fairground ride

• energy transfer of a pendulum passing through equilibrium

why is the energy transfer in a fairground ride ½ m v² = m g h?

• the energy of the fairground ride at max height is GPE. it has no E_k here because it is released from rest

• fairground ride speeds up whilst moving down the track. it passes equilibrium position at Ek_max and GPE_min

• max speed is reached at equilibrium

• energy is conserved, so E_k = loss of GPE from max height

• therefore ½ m v² = m g h

why is the energy transfer in a pendulum bob passing through equilibrium ½ m v² = m g h?

• the energy of a pendulum bob at max height is GPE. it has no E_k here because its either released from rest, or because the pendulum stops momentarily when turning

• pendulum bob speeds up when displaced from equilibrium. it passes equilibrium position at Ek_max and GPE_min

• max speed is reached at equilibrium

• it slows down to reach max height at the other side of the equilibrium position

• energy is conserved, so E_k = loss of GPE from max height

• therefore ½ m v² = m g h

when does a fairground ride have GPE_max?

when at rest on top of the track

when does a fairground ride have GPE_min?

at the bottom of the track

when does a fairground ride have Ek_max?

at the bottom of the track

when does a fairground ride have Ek_min?

when at rest at the top of the track

does the E_k of a fairground ride ever reach zero?

yes

when does the E_k of a fairground ride reach zero?

when the fairground ride is at the top of the track, its at rest so E_k is zero

why is the GPE_max of a fairground ride at the top of the track?

• there is no E_k cuz at this point the fairground ride is at rest

• this is the highest vertical distance of the fairground ride, therefore GPE is at its highest possible value

why is the GPE_min of a fairground ride at the bottom of the track?

• the fairground’s position is not that high off the ground, so it has low GPE

• all the fairground’s GPE has been converted to E_k when the fairground moved down the track

why is the Ek_max of a fairground ride at the bottom of the track?

all the GPE has been transferred to E_k as the ride went down the tracks

why is the Ek_min of a fairground ride at the top of the track?

• at this point, the fairground ride is at rest so there is no E_k

• all the fairground ride’s energy is stored as GPE at this point

what is the energy of a fairground ride at max height?

• GPE_max

• Ek_min

what is the energy of a fairground ride at min height?

• GPE_min

• Ek_max

what is the work done in a fairground ride?

• m g h - ½ m v²

• work done to overcome friction and air resistance

m g h - ½ m v²

the work done in a fairground ride to overcome friction and air resistance

why is the work done in a fairground ride m g h - ½ m v²?

what is the energy at different points in a fairground ride’s motion?

here