Kinetic Energy

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A 3 kg object speeds up from 4 m/s to 10 m/s. Change in kinetic energy?

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90 J

126 J

168 J

210 J

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Subject: physics
Chapter: work, energy and power
Topic: kinetic energy
Difficulty: Easy
Year: 2025

Solution

Correct Answer:

126 J

Kinetic Energy Formula: $K = \frac{1}{2} m v^{2}$ .
Change Calculation: $Δ K = K_{f ina l} - K_{ini t ia l} = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2}$ .
Data Substitution:
- Mass $m = 3$ kg
- $v_{i} = 4$ m/s $⟹ v_{i}^{2} = 16$
- $v_{f} = 10$ m/s $⟹ v_{f}^{2} = 100$
Calculate:
- $Δ K = \frac{1}{2} \times 3 \times (100 - 16)$
- $Δ K = 1.5 \times 84 = 126$ J.
Result: The object gained $126$ J of kinetic energy.

This easy difficulty physics question is from the chapter work, energy and power, covering the topic of kinetic energy. It appeared in the 2025 exam.

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