Recall and use the equation for kinetic energy E_k = ½ m v^2

1.7.1 Energy – Kinetic Energy

Learning Objective

Recall and use the equation for kinetic energy:

\$E_k = \frac{1}{2} m v^2\$

Key Concepts

• Kinetic energy (E_k) is the energy possessed by an object because of its motion.
• It depends on the object's mass (m) and its speed (v).
• The SI unit of kinetic energy is the joule (J), where 1 J = 1 kg·m²·s⁻².

Formula Derivation (Brief)

The work done to accelerate an object from rest to speed v is the integral of the force over the distance travelled:

\$W = \int0^s F \, ds = \int0^v m a \, \frac{v}{a} \, dv = \int_0^v m v \, dv = \frac{1}{2} m v^2\$

Since the work done on the object becomes its kinetic energy, we obtain the kinetic‑energy formula.

Variables and Units

Worked Example

Problem: A 0.15 kg tennis ball is struck and leaves the racket with a speed of 35 m s⁻¹. Calculate its kinetic energy.

Solution:

\$E_k = \frac{1}{2} \times 0.15 \times (35)^2\$

\$E_k = 0.075 \times 1225\$

\$E_k = 91.9\ \text{J}\$

Therefore, the tennis ball has a kinetic energy of approximately 92 J.

Common Mistakes to Avoid

• Using velocity squared incorrectly (e.g., forgetting to square the speed).
• Mixing units (e.g., using grams instead of kilograms).
• Confusing kinetic energy with momentum; they are related but not the same.
• Omitting the factor ½ in the formula.

Practice Questions

1. A 2 kg cart moves at 4 m s⁻¹. What is its kinetic energy?
2. A cyclist of mass 70 kg (including bike) travels at 6 m s⁻¹. Calculate the kinetic energy.
3. If a ball’s kinetic energy is 20 J and its mass is 0.2 kg, what is its speed?
4. Two objects have the same kinetic energy. One has a mass of 0.5 kg; the other has a mass of 2 kg. Which one is moving faster, and by what factor?

Answers

1. \$E_k = \frac{1}{2} \times 2 \times 4^2 = 1 \times 16 = 16\ \text{J}\$
2. \$E_k = \frac{1}{2} \times 70 \times 6^2 = 35 \times 36 = 1260\ \text{J}\$
3. Rearrange \$Ek = \frac{1}{2} m v^2\$ → \$v = \sqrt{\frac{2Ek}{m}} = \sqrt{\frac{2 \times 20}{0.2}} = \sqrt{200} \approx 14.1\ \text{m s}^{-1}\$
4. Set \$\frac{1}{2} m1 v1^2 = \frac{1}{2} m2 v2^2\$. Hence \$v1/v2 = \sqrt{m2/m1} = \sqrt{2/0.5}= \sqrt{4}=2\$. The 0.5 kg object moves twice as fast.