Mechanical Energy

Theory Exercises

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 (E_k)

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 (E_g)

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 (E_e)

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 \(E_g = 0\)):
\(\frac{1}{2}mv_f^2 = 525 \implies v_f^2 = \frac{2 \times 525}{2} = 525 \implies \boxed{v_f = \sqrt{525} \approx 22.91\,\text{m/s}}\)