The combined mass of a cyclist and the bike is 90 kg . Determine the increase in kinetic energy when the speed increases from 6.0 km/h to 12 km/h .

Correct answer: C. 375 J Initial speed v_1 = 6.0 km/h = 6.0 × (5)/(18) m/s = (5)/(3) m/s Final speed v_2 = 12 km/h = 12 × (5)/(18) m/s = (10)/(3) m/s Initial kinetic energy K_1 = (1)/(2) m v_1^2 = (1)/(2) × 90 × ((5)/(3))^2 = 45 × (25)/(9) = 125 J Final kinetic energy K_2 = (1)/(2) m v_2^2 = (1)/(2) × 90 × ((10)/(3))^2 = 45 × (100)/(9) = 500 J Increase in kinetic energy Δ K = K_2 - K_1 = 500 - 125 = 375 J

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Work and Energy

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Work and Energy

The combined mass of a cyclist and the bike is 90 kg . Determine...

The combined mass of a cyclist and the bike is

90 \text{ kg}

. Determine the increase in kinetic energy when the speed increases from

6.0 \text{ km/h}

12 \text{ km/h}

Options

A $4860 \text{ J}$
B $750 \text{ J}$
C $375 \text{ J}$
D $125 \text{ J}$

Correct answer

C. $375 \text{ J}$

Step-by-step solution

Initial speed

v_1 = 6.0 \text{ km/h} = 6.0 \times \dfrac{5}{18} \text{ m/s} = \dfrac{5}{3} \text{ m/s}

Final speed

v_2 = 12 \text{ km/h} = 12 \times \dfrac{5}{18} \text{ m/s} = \dfrac{10}{3} \text{ m/s}

Initial kinetic energy

K_1 = \dfrac{1}{2} m v_1^2 = \dfrac{1}{2} \times 90 \times \left(\dfrac{5}{3}\right)^2 = 45 \times \dfrac{25}{9} = 125 \text{ J}

Final kinetic energy

K_2 = \dfrac{1}{2} m v_2^2 = \dfrac{1}{2} \times 90 \times \left(\dfrac{10}{3}\right)^2 = 45 \times \dfrac{100}{9} = 500 \text{ J}

Increase in kinetic energy

\Delta K = K_2 - K_1 = 500 - 125 = 375 \text{ J}

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