Mechanical Energy

Definition

Mechanical energy is the sum of kinetic energy and potential energy in a system.

$$ E_m = E_k + E_g = \frac{1}{2}mv^2 + mgh $$

Where:

  • $E_k$ = kinetic energy (energy of motion)
  • $E_g$ = potential energy (stored energy due to position)

Types of Mechanical Energy

1. Kinetic Energy (Ek)

Energy of motion:

$$ E_k = \frac{1}{2}mv^2 $$

Characteristics:

  • Zero when object is at rest
  • Increases with velocity squared
  • Always positive

2. Gravitational Potential Energy (Eg)

Energy stored due to height:

$$ E_g = mgh $$

Characteristics:

  • Relative to chosen reference level
  • Higher position = more PE
  • Zero at reference level

3. Elastic Potential Energy (Ee)

Energy stored in compressed/stretched materials:

$$ E_e = \frac{1}{2}kx^2 $$

Conservation of Mechanical Energy

The Principle

In the absence of friction and other dissipative forces, the total mechanical energy of a system remains constant.

$$ E_{\text{total}} = E_k + E_g = \text{constant} $$

Mathematical Expression

At any two points in motion:

$$ E_{k1} + E_{g1} = E_{k2} + E_{g2} $$
$$ \frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2 $$

Solved Exercises

Exercise 1. A 3 kg rock is dropped from a cliff 45 m high. No air resistance, g = 10 m/s².

Find: final velocity before impact.

$v^2 = 2gh = 2(10)(45) = 900 \implies \boxed{v = 30\,\text{m/s}}$

Exercise 2. A vehicle's engine produces 120 000 J with an efficiency of 25%. The vehicle has a mass of 100 kg.

Find: (a) available energy, (b) maximum speed.

(a) Available (useful) energy:

$E_{\text{useful}} = \eta \cdot E_{\text{total}} = 0.25 \times 120\,000 \implies \boxed{E_{\text{useful}} = 30\,000\,\text{J}}$

(b) Maximum speed — all useful energy converts to kinetic energy:

$E_{\text{useful}} = \frac{1}{2}mv^2 \implies 30\,000 = \frac{1}{2}(100)v^2$

$v^2 = 600 \implies \boxed{v = \sqrt{600} \approx 24.49\,\text{m/s}}$

Exercise 3. A 2 kg ball is thrown upward with a speed of 14 m/s.

Find: initial kinetic energy.

$E_k = \frac{1}{2}mv^2 = \frac{1}{2}(2)(14)^2 = \frac{1}{2}(2)(196) \implies \boxed{E_k = 196\,\text{J}}$

Exercise 4. A 5 kg object falls from a height of 20 m. No air resistance, g = 10 m/s².

Find: speed just before hitting the ground.

All $E_g$ converts to $E_k$:

$mgh = \frac{1}{2}mv^2 \implies v^2 = 2gh = 2(10)(20) = 400 \implies \boxed{v = 20\,\text{m/s}}$

Exercise 5. A ball is thrown vertically upward from a height of 2 m with an initial speed of 20 m/s. No air resistance, g = 10 m/s².

Find: (a) maximum height, (b) speed when it hits the ground.

Known data: $h_0 = 2\,\text{m}$, $v_0 = 20\,\text{m/s}$, $g = 10\,\text{m/s}^2$

(a) Maximum height — at the top, $v = 0$, so $E_k = 0$. Apply conservation:

$E_{k0} + E_{g0} = E_{g\,\text{max}}$

$\frac{1}{2}mv_0^2 + mgh_0 = mgh_{\text{max}}$

$\frac{1}{2}(20)^2 + 10 \cdot 2 = 10 \cdot h_{\text{max}}$

$200 + 20 = 10\,h_{\text{max}} \implies \boxed{h_{\text{max}} = 22\,\text{m}}$

(b) Speed at ground level — set $h = 0$, so $E_g = 0$:

$\frac{1}{2}mv_0^2 + mgh_0 = \frac{1}{2}mv_f^2$

$200 + 20 = \frac{1}{2}v_f^2 \implies v_f^2 = 440 \implies \boxed{v_f = \sqrt{440} \approx 20.98\,\text{m/s}}$

Exercise 6. A roller coaster car of mass 400 kg starts from rest at a height of 30 m. No friction, g = 10 m/s².

Find: speed at the bottom of the track.

$mgh = \frac{1}{2}mv^2 \implies v^2 = 2(10)(30) = 600 \implies \boxed{v = \sqrt{600} \approx 24.49\,\text{m/s}}$

Exercise 7. A cyclist of mass 75 kg rides at 10 m/s.

Find: kinetic energy.

$E_k = \frac{1}{2}mv^2 = \frac{1}{2}(75)(10)^2 = \frac{1}{2}(75)(100) \implies \boxed{E_k = 3750\,\text{J}}$

Exercise 8. A 1.5 kg object has 300 J of kinetic energy.

Find: velocity.

$E_k = \frac{1}{2}mv^2 \implies 300 = \frac{1}{2}(1.5)v^2 \implies v^2 = \frac{600}{1.5} = 400 \implies \boxed{v = 20\,\text{m/s}}$

Exercise 9. A 2 kg stone is thrown downward from a 25 m building with an initial velocity of 5 m/s. No air resistance, g = 10 m/s².

Find: velocity just before reaching the ground.

Total mechanical energy at the top:

$E_m = \frac{1}{2}mv_0^2 + mgh = \frac{1}{2}(2)(5)^2 + (2)(10)(25) = 25 + 500 = 525\,\text{J}$

At ground level ($h = 0$, so $E_g = 0$):

$\frac{1}{2}mv_f^2 = 525 \implies v_f^2 = \frac{2 \times 525}{2} = 525 \implies \boxed{v_f = \sqrt{525} \approx 22.91\,\text{m/s}}$

1. What is mechanical energy? (1 points)

2. In the absence of friction, what happens to total mechanical energy? (1 points)

3. For a pendulum at its highest point: (1 points)

4. For a pendulum at its lowest point: (1 points)

5. A ball is dropped from rest at height h. Which statement is true (no friction)? (1 points)

6. What happens to mechanical energy when friction is present? (1 points)

7. For a projectile launched at 45° angle, at maximum height: (1 points)

8. A roller coaster at the bottom of a dip (no friction) has: (1 points)

9. Energy is dissipated by friction as: (1 points)

10. For conservation of mechanical energy equation E₁ = E₂, what must be true? (1 points)

11. Mechanical energy calculations: (4 points)

a ) A 2 kg object at height 5 m, at rest. Mechanical energy = _____ J (g = 10 m/s², reference at ground)

b ) A 1 kg ball thrown upward at 20 m/s from ground. Total mechanical energy = _____ J (g = 10 m/s²)

c ) A 3 kg object has E<sub>k</sub> = 45 J and E<sub>g</sub> = 75 J. Mechanical energy = _____ J

d ) A ball at height h has E<sub>g</sub> = 200 J and total mechanical energy = 250 J. E<sub>k</sub> = _____ J

12. Energy conservation scenarios: (2 points)

a ) A 4 kg object dropped from 10 m reaches ground at what speed? v = _____ m/s (no friction, g = 10 m/s²)

b ) Pendulum at 1 m height has E<sub>k</sub> = 40 J. Maximum height reached = _____ m (mass = 2 kg, g = 10 m/s²)

13. A ball is thrown vertically upward from a height of 2 m above the ground with an initial speed of 20 m/s (g = 10 m/s², no air resistance): (2 points)

a ) Maximum height reached by the ball = _____ m

b ) Velocity when the ball hits the ground = _____ m/s (round to 2 decimal places)

14. A vehicle with an internal combustion engine produces 120 000 J with an efficiency of 25% and a mass of 100 kg: (2 points)

a ) Available (useful) energy = _____ J

b ) Maximum velocity the vehicle could reach with that energy = _____ m/s (round to 2 decimal places)

15. Basic kinetic and potential energy calculations (g = 10 m/s²): (6 points)

a ) A 2 kg ball thrown upward at 14 m/s. Initial kinetic energy = _____ J

b ) A 5 kg object falls from 20 m. Speed just before hitting the ground = _____ m/s

c ) A 3 kg rock dropped from 45 m. Final velocity before impact = _____ m/s

d ) A roller coaster car (400 kg) starts from rest at 30 m. Speed at the bottom = _____ m/s (round to 2 decimal places)

e ) A cyclist (75 kg) rides at 10 m/s. Kinetic energy = _____ J

f ) A 1.5 kg object has 300 J of kinetic energy. Its velocity = _____ m/s

16. A 2 kg stone is thrown downward from a 25 m building with an initial velocity of 5 m/s (g = 10 m/s², no air resistance): (1 point)

a ) Velocity just before reaching the ground = _____ m/s (round to 2 decimal places)