5.3 Use gravitational and elastic potential energy | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Work, Energy, and Power

EN PH 131 Dynamics · Study topic 5.3

Potential energy tracks energy associated with position or configuration. Near Earth’s surface, an object gains gravitational potential energy as it rises. An ideal spring stores elastic potential energy when extended or compressed from its unstretched length. These changes can connect initial and final states without finding the entire force and acceleration history. In each example, we define the system and observer, choose a reference and positive direction, and check that the assumptions make an energy method suitable.

What you will learn

Define gravitational and elastic potential energy and identify their reference choices.
Use work–energy to relate changes in potential energy to changes in kinetic energy.
Apply energy methods to motion under gravity, an ideal spring, or both.
Check energy calculations using signs, units, and limiting cases.

1. Gravitational potential energy and reference choices

The system is the object or objects whose energy we track. The observer describes the motion, and the reference frame is the coordinate system used to describe positions and velocities. In the examples, the observer is fixed to the ground. Use horizontal and vertical axes, with upward positive on the vertical axis.

Near Earth’s surface, a particle of mass

m

has weight of magnitude

mg

, directed downward. Here

g

is gravitational acceleration. If the particle’s vertical coordinate is

y

, its gravitational potential energy can be written

V_g=mgy+C

. The constant

C

depends on the chosen zero level. Only changes in potential energy affect the energy balance.

Between heights

y_1

and

y_2

, the change in gravitational potential energy is

mg(y_2-y_1)

. A rise makes this change positive; a drop makes it negative. This matches gravity’s work: gravity does negative work during an upward displacement and positive work during a downward displacement. Weight is a force measured in newtons; mass is measured in kilograms.

V_g=mgy+C,\qquad \Delta V_g=mg(y_2-y_1)

Choose a convenient height as the zero of gravitational potential energy and use it consistently.
With upward position positive, a rise increases gravitational potential energy.
A potential-energy value can be negative if the chosen zero level makes it so; the energy difference is what matters.

2. Elastic potential energy and the work–energy rule

An ideal linear spring exerts a force that points toward its unstretched state. Let

x

be its signed deformation measured from that state along the spring axis. Extension and compression have opposite signs. The spring force is

F_s=-kx

, where

k

is stiffness, measured in newtons per metre. The minus sign indicates that the force opposes the deformation.

The elastic potential energy, measured relative to the unstretched state, is

V_e=\tfrac12kx^2

. Since the deformation is squared, this stored energy is nonnegative for either compression or extension. Its change between deformations

x_1

and

x_2

is

\tfrac12k(x_2^2-x_1^2)

. It is positive when the spring stores more energy in the final state and negative when it stores less.

Kinetic energy for a particle of mass

m

and speed

v

is

T=\tfrac12mv^2

. If gravity and ideal spring forces are the only forces doing work, the sum of kinetic and relevant potential energies is the same in the initial and final states. The reason is that the work of each of these forces equals the negative change in its potential energy.

If other forces do work, include their work in the balance instead of claiming mechanical energy is conserved. For example,

U_{1\to2}

can represent the work of forces not represented by the potential-energy terms. Identify those forces and their work before applying the equation. Energy equations generally determine speed, not direction; motion information or the geometry supplies direction.

T_1+V_1+U_{1\to2}=T_2+V_2

Measure spring deformation from its unstretched length, not from an arbitrary coordinate origin.
Spring force changes direction across the unstretched state; spring potential energy does not become negative.
Mechanical energy is conserved only when the work of forces outside the chosen potential-energy terms is zero.

3. Set up, solve, and check

Begin by identifying the system, the ground-fixed observer and frame, and the initial and final states. Set coordinates and positive directions. State the gravitational zero-height reference and, for each spring, the unstretched position at which deformation is zero. A small sketch can clarify which way is up or how the spring is deformed.

List known values—such as mass, spring stiffness, initial speed, height change, and spring deformations—and identify the requested speed or position. Choose the work–energy relation when the configurations and energy changes are known and the requested quantity is a speed or position. State whether friction or any other force does work.

Solve symbolically before substituting numbers. Every energy term must have units of joules:

1\,\mathrm{J}=1\,\mathrm{N\,m}=1\,\mathrm{kg\,m^2/s^2}

. A positive square root gives speed magnitude. Use the stated motion to report direction. Check that zero spring deformation gives zero elastic potential energy, a rise increases gravitational potential energy, and the final state is reachable with the available energy.

Use the same references and coordinate choices for both states.
A negative change in potential energy can simply mean stored potential energy decreased.
Check dimensions, signs, direction, initial conditions, and whether the energy balance permits the final state.

Worked example

A particle descending a ramp

A cart is modelled as a particle of mass

2.0\,\mathrm{kg}

. It starts from rest and descends a smooth ramp, losing

1.5\,\mathrm{m}

of vertical height. Find its speed at the lower point. Neglect friction and air resistance.

Cart and gravity

Gravity acts downward; the smooth ramp’s normal force does no work.

Define the states and reference
Take the cart as the system and use a ground-fixed observer. Let vertical position be positive upward, and choose the lower point as zero gravitational potential energy. The initial state is at rest, $1.5\,\mathrm{m}$ above the final state.
Apply energy conservation
Gravity does work, but it is represented by gravitational potential energy. The ramp’s normal force is perpendicular to the motion and does no work. With no friction or air resistance, mechanical energy is conserved.
$mg(1.5\,\mathrm{m})=\frac12mv_2^2$
Find the speed
Cancel the mass and use $g=9.81\,\mathrm{m/s^2}$ . The cart moves toward the lower point, while the energy equation gives the speed magnitude.
$v_2=\sqrt{2g(1.5\,\mathrm{m})}=5.42\,\mathrm{m/s}$

Answer: The cart’s speed at the lower point is

5.42\,\mathrm{m/s}

.

Check: Inside the square root, the units are

(\mathrm{m/s^2})(\mathrm{m})=\mathrm{m^2/s^2}

, so the result has units of speed. A greater vertical drop would give a greater speed.

Worked example

A spring launches a block

A

0.80\,\mathrm{kg}

block is released from rest on a horizontal, frictionless track. It is pressed against an ideal spring of stiffness

200\,\mathrm{N/m}

, compressing it by

0.12\,\mathrm{m}

from its unstretched length. Find the block’s speed as the spring returns to its unstretched length.

Block and spring force

During release from compression, the spring force points toward the unstretched position.

Define system and states
Take the block and ideal spring as the energy system, observed from the ground. Let the spring axis be positive to the right. The initial deformation is $x_1=-0.12\,\mathrm{m}$ , and the final deformation is $x_2=0$ . Initially the block is at rest.
Relate spring energy to kinetic energy
The track is horizontal, so gravitational potential energy does not change. With no friction, the spring’s lost stored energy becomes the block’s kinetic energy.
$\frac12kx_1^2=\frac12mv_2^2+\frac12kx_2^2$
Calculate
Substitute the given stiffness, mass, and deformations. As the spring expands from compression, the block moves to the right.
$v_2=\sqrt{\frac{k}{m}(x_1^2-x_2^2)}=1.90\,\mathrm{m/s}$

Answer: The block’s speed at the unstretched position is

1.90\,\mathrm{m/s}

, directed to the right.

Check: The units of

kx^2/m

reduce to

\mathrm{m^2/s^2}

, as required under the square root. If the initial compression were zero, the available spring energy and predicted speed would both be zero.

Worked example

A spring lifts a collar

A

1.2\,\mathrm{kg}

collar is released from rest while a vertical spring is compressed by

0.20\,\mathrm{m}

. The collar rises

0.10\,\mathrm{m}

before the spring reaches its unstretched length. The spring stiffness is

300\,\mathrm{N/m}

. With no friction, determine the collar’s speed at that instant.

Collar and vertical spring

At the initial compressed state, the spring pushes upward against gravity.

Set references and states
Take the collar and ideal spring as the system, observed from a ground-fixed frame. Upward is positive, and the initial height is the gravitational reference. Initially the collar is at rest and the spring deformation magnitude is $0.20\,\mathrm{m}$ . Finally, the collar has risen $0.10\,\mathrm{m}$ and the spring is unstretched.
Write the energy balance
The spring’s initial stored energy becomes gravitational potential energy and kinetic energy. Since there is no friction, mechanical energy is conserved.
$\frac12k(0.20\,\mathrm{m})^2=mg(0.10\,\mathrm{m})+\frac12mv_2^2$
Solve and interpret
The spring initially stores $6.00\,\mathrm{J}$ . The rise requires about $1.18\,\mathrm{J}$ , leaving positive energy for motion. The collar is rising at the instant described.
$v_2=\sqrt{\frac{k(0.20\,\mathrm{m})^2-2mg(0.10\,\mathrm{m})}{m}}=2.83\,\mathrm{m/s}$

Answer: The collar’s speed when the spring reaches its unstretched length is

2.83\,\mathrm{m/s}

, upward.

Check: The radicand is about

8.04\,\mathrm{m^2/s^2}

and is positive, so the stated final position is reachable. The units reduce to speed squared. If the initial spring energy were less than the required increase in gravitational potential energy, the collar could not reach that position from rest.

Common mistakes and how to avoid them

Treating gravitational potential energy as always positive.

Correction: Its value depends on the chosen zero level. Use the same reference in both states; a negative value does not change the physical energy difference.

Using signed spring deformation directly as elastic potential energy.

Correction: Use

V_e=\tfrac12kx^2

. Both compression and extension store nonnegative energy.

Assuming mechanical energy is conserved when friction does work.

Correction: Include work by forces not represented by potential-energy terms, or do not claim mechanical energy is conserved.

Reporting direction from a positive square root alone.

Correction: The energy equation generally gives speed magnitude. Use the stated motion and coordinate information to identify direction.

Lesson summary

Near Earth, gravitational potential-energy change equals $mg$ times the vertical rise.
An ideal spring stores $\tfrac12kx^2$ relative to its unstretched length.
Use the work–energy balance and identify any work not represented by potential-energy terms.
Check references, signs, SI units, direction, and whether the final state is reachable.

Check your understanding

Question 1

A particle rises by

0.40\,\mathrm{m}

. What is the change in gravitational potential energy if its mass is

3.0\,\mathrm{kg}

?

$-11.8\,\mathrm{J}$
$+11.8\,\mathrm{J}$
$+7.36\,\mathrm{J}$
correctIndex: 1, "explanation": "With upward positive, $\Delta V_g=mg\Delta y=(3.0)(9.81)(0.40)=11.8\,\mathrm{J}$ , a positive change."

Show answer and explanation

+11.8\,\mathrm{J}

With upward positive,

\Delta V_g=mg\Delta y=(3.0)(9.81)(0.40)=11.8\,\mathrm{J}

, a positive change.

Question 2

An ideal spring with

k=100\,\mathrm{N/m}

is compressed by

0.15\,\mathrm{m}

. How much elastic potential energy is stored?

$1.13\,\mathrm{J}$
$-1.13\,\mathrm{J}$
$15.0\,\mathrm{J}$
correctIndex: 0, "explanation": "The stored energy is $\tfrac12kx^2=\tfrac12(100)(0.15)^2=1.125\,\mathrm{J}$ , which rounds to $1.13\,\mathrm{J}$ ."

Show answer and explanation

1.13\,\mathrm{J}

The stored energy is

\tfrac12kx^2=\tfrac12(100)(0.15)^2=1.125\,\mathrm{J}

, which rounds to

1.13\,\mathrm{J}

.

Question 3

With no friction, a block moves on a horizontal track while an ideal spring relaxes. Which statement is correct?

The spring’s elastic potential energy decreases as the block’s kinetic energy increases.
The spring’s elastic potential energy and the block’s kinetic energy both decrease.
The spring’s elastic potential energy increases as the block’s kinetic energy increases.
correctIndex: 0, "explanation": "On a horizontal frictionless track, gravitational potential energy is unchanged. The spring’s lost stored energy becomes kinetic energy."

Show answer and explanation

The spring’s elastic potential energy decreases as the block’s kinetic energy increases.

On a horizontal frictionless track, gravitational potential energy is unchanged. The spring’s lost stored energy becomes kinetic energy.

Key terms

Gravitational potential energy: Energy associated with position in a gravitational field; near Earth its change is $mg$ times vertical displacement.
Elastic potential energy: Energy stored in an ideal spring due to deformation from its unstretched length, equal to $\tfrac12kx^2$ .
Reference level: A chosen position assigned zero potential energy; changing it shifts energy values but not energy differences.
Mechanical energy: The sum of kinetic energy and the gravitational and elastic potential energies included in the model.

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5.3 · Use gravitational and elastic potential energy

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 Dynamics · Study topic 5.3

What you will learn

1. Gravitational potential energy and reference choices

2. Elastic potential energy and the work–energy rule

3. Set up, solve, and check

A particle descending a ramp

A spring launches a block

A spring lifts a collar

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through EN PH 131

About this lesson and its review