Question

The speed of a body of mass 10 kg changes from 20 ms$^{-1}$ to 30 ms$^{-1}$. The increase in the kinetic energy of the body is

• A. 1250 J
• B. 4500 J
• C. 2500 J
• D. 3000 J

Hint:

To find the increase in kinetic energy, calculate the initial and final kinetic energies using \( KE = \frac{1}{2}mv^2 \), then subtract: \( \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \).

Answer 1

The Correct Option is C

The kinetic energy of a body is given by: \[ KE = \frac{1}{2}mv^2 \] Where:
- \( m = 10 \, \text{kg} \) (mass of the body),
- \( v \) is the speed of the body.
Initial kinetic energy (\( v_1 = 20 \, \text{ms}^{-1} \)): \[ KE_1 = \frac{1}{2} \times 10 \times (20)^2 = \frac{1}{2} \times 10 \times 400 = 2000 \, \text{J} \] Final kinetic energy (\( v_2 = 30 \, \text{ms}^{-1} \)): \[ KE_2 = \frac{1}{2} \times 10 \times (30)^2 = \frac{1}{2} \times 10 \times 900 = 4500 \, \text{J} \] The increase in kinetic energy is: \[ \Delta KE = KE_2 - KE_1 = 4500 - 2000 = 2500 \, \text{J} \] So, the increase in the kinetic energy of the body is 2500 J.

Answer 2

Given:
Mass of the body, \( m = 10\, \text{kg} \)
Initial speed, \( u = 20\, \text{ms}^{-1} \)
Final speed, \( v = 30\, \text{ms}^{-1} \)

Initial kinetic energy,
\[ KE_i = \frac{1}{2} m u^2 = \frac{1}{2} \times 10 \times (20)^2 = 5 \times 400 = 2000\, \text{J} \]

Final kinetic energy,
\[ KE_f = \frac{1}{2} m v^2 = \frac{1}{2} \times 10 \times (30)^2 = 5 \times 900 = 4500\, \text{J} \]

Increase in kinetic energy,
\[ \Delta KE = KE_f - KE_i = 4500 - 2000 = 2500\, \text{J} \]

Final answer:
\[ \boxed{2500\, \text{J}} \]