Physics Work & Energy SZABMU 2022
PMDC Verified Question 48 of 108
A car of mass 800 kg accelerates from \( 20 \text{ ms}^{-1} \) to \( 30 \text{ ms}^{-1} \), the increase in K.E will be:
A
2 J
B
200 kJ
C
200 J
D
2 kJ
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: 200 kJ
Concept:

The increase in kinetic energy is simply the final kinetic energy minus the initial kinetic energy.

Formula:

$$ \Delta K.E = \frac{1}{2}m(v_f^2 - v_i^2) $$

Solution:

  • Mass (\( m \)) = 800 kg


  • Initial velocity (\( v_i \)) = \( 20 \text{ m/s} \)


  • Final velocity (\( v_f \)) = \( 30 \text{ m/s} \)


  • \( \Delta K.E = \frac{1}{2}(800)(30^2 - 20^2) \)


  • \( \Delta K.E = 400(900 - 400) = 400(500) \)


  • \( \Delta K.E = 200,000 \text{ J} \)


  • Converting to kilojoules: \( 200,000 \text{ J} = 200 \text{ kJ} \).


Why other options are incorrect:

A common mistake is doing \( \frac{1}{2}m(v_f - v_i)^2 \), which yields \( 400(10)^2 = 40,000 \text{ J} = 40 \text{ kJ} \). Failing to convert Joules to kJ leads to magnitude errors.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

Previous Question All Work & Energy MCQs Next Question

Want to solve full-length papers under timed exam conditions?

Practice with zero-scroll lockdown sprints, dynamic latency zone timers, live peer selection telemetry, and the automated Amber mistake recovery loop.

Enter Arena Sprint iOS App Android App