Energy and power of a falling mass

SPH3U Grade 11 Physics · Energy and Society

For a work-site safety demonstration, a physics class gathers around a scaffold erected in the school parking lot. A claw hammer, weighed beforehand on the shop balance at 600 g, rests at the edge of a platform whose deck a tape measure puts 4.50 m above the ground. A boot catches the handle, the hammer tips over the edge, and it falls freely — nothing snags it on the way down — until it thuds into the gravel below. A second student films the drop against a metre-stick backdrop, and counting frames on the video puts the fall at 0.958 s. Find the gravitational potential energy the hammer had on the platform, the speed it reaches at the ground, and the average power gravity delivers during the fall.

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given

m = 600 g — Mass of the hammer
h = 4.5 m — Height of the platform
t = 0.958 s — Duration of the free fall

Determine

(a)the gravitational potential energy of the hammer on the platform
(b)the speed it reaches at the ground
(c)the average power gravity delivers during the fall

Step 1 of 3(a) · 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

Your values, in your units

U = \left( 600\ \text{g} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \left( 4.5\ \text{m} \right)

Converted to base units

U = \left( 0.6\ \text{kg} \right) \, \left( 9.80665\ \text{m/s}^{2} \right) \, \left( 4.5\ \text{m} \right)

Answer

U = 26.478\ \text{J}

Carried onward at full precision, not this rounded figure.

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

Step 2 of 3(b) · 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}}

Eₖ26.478 Jcarried from step 1

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 = 9.3947\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Kinetic Energy solver →

Step 3 of 3(c) · solve for Power

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

Rearranged for P

P = \frac{W}{t}

W26.478 Jcarried from step 1

Your values, in your units

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

Answer

P = 27.639\ \text{W}

Carried onward at full precision, not this rounded figure.

Open the Power (P = W/t) solver →

Answer

Therefore the hammer holds 26.5 J of gravitational potential energy on the platform, arrives at the ground at 9.39 m/s, and gravity delivers those joules at an average 27.6 W across the 0.958 s of the fall.

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.