Mechanical Energy

Table of Contents
1. Potential energy
2. Kinetic energy
3. Mechanical energy
4. Conservation of mechanical energy
5. Exercises
6. Summary
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All objects have energy. The word energy comes from the Greek energeia, meaning activity or operation. Energy is closely linked to mass and, in an isolated system, cannot be created or destroyed. This chapter considers gravitational potential energy, kinetic energy and their sum, the mechanical energy. The focus is on definitions, formulae, worked examples and exercises suitable for a school-level treatment.

Potential energy

Potential energy is the energy an object has because of its position or state. There are several kinds of potential energy (gravitational, chemical, electrical, etc.); here we consider gravitational potential energy only.

Gravitational potential energy is the energy an object has due to its position in a gravitational field relative to some chosen reference point. The symbol used here is E_P.

The formula for gravitational potential energy near the Earth's surface is

E_P = m g h

• m is the mass (kg)
• g is the gravitational acceleration (take g = 9.8 m·s^-2)
• h is the perpendicular height above the chosen reference point (m)

Example calculation: a 1 kg suitcase placed on top of a cupboard 2 m above the floor. Its gravitational potential energy relative to the floor is

E_P = m g h = (1 kg)(9.8 m·s^-2)(2 m) = 19.6 J.

If the suitcase falls, its potential energy decreases with height. Halfway down (h = 1.0 m) the potential energy is 9.8 J, and at floor level (h = 0) it is 0 J. Thus potential energy is maximum at maximum height and decreases as the object falls.

Example 1: Gravitational potential energy

SOLUTION

The mass is m = 1 kg and the height is h = 4 m. Using E_P = m g h with g = 9.8 m·s^-2:

E_P = (1 kg)(9.8 m·s^-2)(4 m) = 39.2 J at the top.

On the ground (h = 0), E_P = 0 J.

Example 2: Netball

SOLUTION

Mass m = 0.50 kg. g = 9.8 m·s^-2.

Height of ball above ground when about to shoot: h = 1.7 + 0.50 = 2.2 m. So

E_P = (0.50 kg)(9.8 m·s^-2)(2.2 m) = 10.78 J.

At the net (h = 2.5 m):

E_P = (0.50 kg)(9.8 m·s^-2)(2.5 m) = 12.25 J.

On the ground (h = 0): E_P = 0 J.

Kinetic energy

Kinetic energy is the energy an object has due to its motion. The symbol used here is E_K. The formula is

E_K = 1/2 m v²

• m is the mass (kg)
• v is the speed or velocity magnitude (m·s^-1)

Kinetic energy depends on both mass and the square of the velocity. Doubling the speed quadruples the kinetic energy; doubling the mass doubles the kinetic energy.

Example: the 1 kg suitcase at the top of the cupboard is at rest, so E_K = 0. If it reaches the ground with speed 6.26 m·s^-1, then

E_K = 1/2 (1 kg)(6.26 m·s^-1)² = 19.6 J.

Question: Falling brick (kinetic energy)

SOLUTION

At the top the brick is at rest so E_K = 0 J.

At the ground: E_K = 1/2 (1 kg)(8.85 m·s^-1)² = 39.2 J.

Example: Two moving animals

A mother sheep (m = 80 kg) and its lamb (m = 25 kg) both run at v = 2.7 m·s^-1. Their kinetic energies are:

E_{K, sheep} = 1/2 (80 kg)(2.7 m·s^-1)² = 291.6 J.

E_{K, lamb} = 1/2 (25 kg)(2.7 m·s^-1)² = 91.13 J.

Although their speeds are equal, the larger mass gives the sheep the greater kinetic energy.

Checking units

The SI unit for energy is the joule (J). Using fundamental units, kinetic energy has units kg·m²·s^-2, which is equivalent to N·m = J, since 1 N = 1 kg·m·s^-2.

Example: Bullet with mixed units

SOLUTION

Convert mass to kg: 150 g = 0.150 kg. Then

E_K = 1/2 (0.150 kg)(960 m·s^-1)² = 69 120 J.

Mechanical energy

Mechanical energy of a system is the sum of its gravitational potential energy and kinetic energy. Denote mechanical energy by E_M.

E_M = E_P + E_K = m g h + 1/2 m v²

Example: Mechanical energy of a ball

SOLUTION

m = 0.15 kg, h = 2.0 m, E_K = 20 J. Compute potential energy:

E_P = m g h = (0.15 kg)(9.8 m·s^-2)(2.0 m) = 2.94 J.

Total mechanical energy E_M = E_P + E_K = 2.94 J + 20 J = 22.94 J.

Conservation of mechanical energy

Law of Conservation of Energy: Energy cannot be created or destroyed, but may be converted from one form to another.

In a closed system with no dissipative forces (no friction, no air resistance), the Law of Conservation of Mechanical Energy states that the total mechanical energy remains constant. Thus, at any two instants in the motion

E_P1 + E_K1 = E_P2 + E_K2

When using this principle, the path taken by the object is irrelevant; only the heights and speeds at the chosen instants matter.

Example: Suitcase falling from cupboard (conservation)

At the top: E_M1 = m g h + 0. At the bottom: E_M2 = 0 + 1/2 m v². Setting E_M1 = E_M2 for m = 1 kg, h = 2.0 m:

(1 kg)(9.8 m·s^-2)(2.0 m) = 1/2 (1 kg) v²

19.6 J = 1/2 v² ⇒ v² = 39.2 m²·s^-2 ⇒ v = 6.26 m·s^-1.

Thus the potential energy lost is converted into kinetic energy (ignoring dissipative effects).

Activity: Marble on a pipe (conversion)

Materials: a plastic pipe (≈20 mm diameter), a marble, masking tape and a measuring tape.

Procedure: Fix one end of the pipe to a table so it is parallel to the tabletop. Hold the other end at a steady height above the table and measure the vertical height. Release the marble from the top opening and let it roll through the pipe onto the table. Repeat with the pipe held higher.

Observations and questions:

• When first placed at the top (before release) the marble has negligible speed so its kinetic energy is near zero and its gravitational potential energy equals m g h (with h measured from the tabletop).
• At the exit, the marble has converted that potential energy into kinetic energy and moves faster when the start height was larger.

Conclusion: Ignoring friction, a larger starting height gives greater potential energy and hence a larger final kinetic energy and speed. The total mechanical energy depends only on height and speed, not on the length of the pipe.

Example: Tree trunk falling

SOLUTION

m = 100 kg, h = 5.0 m, g = 9.8 m·s^-2.

Potential energy at top: E_P = m g h = (100 kg)(9.8 m·s^-2)(5 m) = 4900 J.

In the absence of air resistance, all this potential energy becomes kinetic energy at the bottom, so E_K = 4900 J.

Use E_K = 1/2 m v² to find v:

4900 J = 1/2 (100 kg) v² ⇒ 4900 = 50 v² ⇒ v² = 98 ⇒ v = 9.90 m·s^-1.

Example: Pendulum

SOLUTION

Let the change in height be h = 0.50 m. Using conservation of mechanical energy and v_A = 0:

m g h = 1/2 m v².

Cancel m: g h = 1/2 v² ⇒ v² = 2 g h. The mass has cancelled, proving velocity is independent of mass.

Compute v: v = sqrt(2 g h) = sqrt(2 × 9.8 m·s^-2 × 0.50 m) = 3.13 m·s^-1.

Example: Roller coaster

SOLUTION

Use conservation of mechanical energy between start (h₁ = 50 m, v₁ = 0) and top of loop (h₂ = 20 m, v₂):

m g h₁ = 1/2 m v₂² + m g h₂ ⇒ g(h₁ - h₂) = 1/2 v₂².

v₂ = sqrt(2 g (h₁ - h₂)) = sqrt(2 × 9.8 × (50 - 20)) = 24.25 m·s^-1.

Between start and ground (h = 0): m g h₁ = 1/2 m v₃² ⇒ v₃ = sqrt(2 g h₁) = sqrt(2 × 9.8 × 50) = 31.30 m·s^-1.

Example: Inclined plane and climber

SOLUTION

The bottle: h = 10 m, v_initial = 0. Use conservation of mechanical energy:

m g h = 1/2 m v² ⇒ v = sqrt(2 g h) = sqrt(2 × 9.8 × 10) = 14.0 m·s^-1.

The distance travelled along the slope (100 m) is irrelevant to the energy calculation; only the vertical height matters.

Climber: m = 60 kg, change in potential energy descending 10 m is ΔE_P = m g Δh = (60 kg)(9.8 m·s^-2)(10 m) = 5880 J (this is the loss of potential energy).

Exercises

1. Describe the relationship between an object's gravitational potential energy and its:

   1. mass;
   2. height above a reference point.

2. A boy of mass 30 kg climbs onto the roof of a garage. The roof is 2.5 m above the ground.

   1. How much potential energy did the boy gain by climbing onto the roof?
   2. If the boy jumps down, what is the potential energy when he is 1.0 m from the ground?
   3. What is the potential energy when he lands on the ground?

3. A hiker of mass 70 kg walks up a mountain and stays at an overnight hut at 800 m above sea level. The second day she walks to a hut at 500 m above sea level. The third day she returns to her starting point at 200 m above sea level.

   1. What is the potential energy of the hiker at the first hut (relative to sea level)?
   2. How much potential energy has the hiker lost during the second day?
   3. How much potential energy did the hiker have when she started (relative to sea level)?
   4. How much potential energy did the hiker have when she returned to her original starting position?

4. Describe the relationship between an object's kinetic energy and its:

   1. mass;
   2. velocity.

5. A stone of mass 0.100 kg is thrown up with initial speed 3.0 m·s^-1. Calculate its kinetic energy:

   1. as it leaves the thrower's hand;
   2. when it reaches its turning point.

6. A car of mass 700 kg travels at 100 km·h^-1. Calculate its kinetic energy.

7. A tennis ball of mass 0.120 kg is dropped from 5 m (ignore air friction):

   1. What is the potential energy after it has fallen 3 m?
   2. What is the velocity when it hits the ground?

8. A ball rolls down a hill of vertical height 15 m (ignore friction):

   1. What is the gravitational potential energy at the top?
   2. What is the velocity at the bottom?

9. A bullet of mass 0.050 kg is shot vertically upward with muzzle velocity 200 m·s^-1 (ignore air friction). Use energy conservation to determine the maximum height reached.

10. A skier of mass 50 kg starts from rest at the top of a slope of vertical height 6.4 m:

    1. Determine the maximum speed at the bottom if friction is negligible.
    2. Will the skier reach this maximum speed in practice? Explain briefly.

11. A pendulum bob of mass 1.5 kg swings from height A to bottom B with speed at B equal to 4.0 m·s^-1. Calculate the height A (ignore air friction).

12. Prove that in free fall (no air resistance) the velocity of an object depends only on the height through which it has fallen and not on its mass.

Summary

• Gravitational potential energy: E_P = m g h.
• Kinetic energy: E_K = 1/2 m v².
• Mechanical energy: E_M = E_P + E_K.
• Unit of energy: joule (J).
• Conservation: In the absence of dissipative forces, total mechanical energy remains constant; potential energy converts to kinetic energy and vice versa.