Energy and power of a falling mass

SPH3U Grade 11 Physics · Energy and Society

A 600 g claw hammer is knocked off a scaffold platform 4.50 m above the ground and falls freely for 0.958 s. Find the gravitational potential energy it had on the platform, the speed it reaches at the ground, and the average power gravity delivers during the fall.

Step 1 of 3 · solve for Potential energy

Start with the energy stored by position. Nothing is moving yet. Hammers are sold by the gram, but mgh wants kilograms, so the 600 g goes in as 0.600 kg before anything is multiplied.

Rearranged for U

U = m g h

m

h

Your values, in your units

U = \left( 600\ \text{g} \right) \, g \, \left( 4.5\ \text{m} \right)

Converted to base units

U = \left( 0.6\ \text{kg} \right) \, g \, \left( 4.5\ \text{m} \right)

Answer

U = 26.5

in

Carried onward at full precision, not this rounded figure.

Open the Gravitational Potential Energy (U = mgh) solver →

Step 2 of 3 · solve for Speed

In free fall every joule of potential energy becomes kinetic energy, so the answer from step 1 is carried straight in as Eₖ. Notice the mass cancels — a 20 kg toolbox would hit at the same speed.

Rearranged for v

v = \sqrt{\tfrac{2 E_k}{m}}

Eₖ26.5 Jcarried from step 1

m

Your values, in your units

v = \sqrt{\tfrac{2 \, \left( 26.478\ \text{J} \right)}{\left( 600\ \text{g} \right)}}

Converted to base units

v = \sqrt{\tfrac{2 \, \left( 26.478\ \text{J} \right)}{\left( 0.6\ \text{kg} \right)}}

Answer

v = 9.39

in

Carried onward at full precision, not this rounded figure.

Open the Kinetic Energy solver →

Step 3 of 3 · solve for Power

Average power is that same energy delivered over the time of the fall.

Rearranged for P

P = \frac{W}{t}

W26.5 Jcarried from step 1

t

Your values, in your units

P = \frac{\left( 26.478\ \text{J} \right)}{\left( 0.958\ \text{s} \right)}

Answer

P = 27.6

in

Carried onward at full precision, not this rounded figure.

Open the Power (P = W/t) solver →

Why this order

This is the standard energy-conservation sequence, and the reason it is taught as a chain rather than three separate calculations is step 2. Students reliably compute the potential energy, then reach for a kinematics equation to find the speed — which works, but misses the point. Conservation of energy gets there in one line, and it keeps working when the path is a curved ramp where the kinematics equations do not apply.

The mass cancelling in step 2 is worth pausing on: mgh = ½mv² gives v = √(2gh) regardless of m. Galileo argued exactly this in Two New Sciences, and the Apollo 15 crew dropped a hammer and a falcon feather on the Moon in 1971 to show it holds when there is no air to spoil it.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.