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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

A car weighing 1400 kg is travelling up a hill at a speed of 54...

A car weighing 1400 kg1400 \text{ kg} is travelling up a hill at a speed of 54 km/h54 \text{ km/h} when its motor stops. If it is just able to reach the destination which is at a height of 10 m10 \text{ m} above the point, determine the work done against friction (the negative of the work done by friction). Assume g=9.8 m/s2g = 9.8 \text{ m/s}^2.

Options

  1. A20300 J20300 \text{ J}
  2. B294700 J294700 \text{ J}
  3. C17500 J17500 \text{ J}
  4. D177800 J177800 \text{ J}

Correct answer

A. 20300 J20300 \text{ J}

Step-by-step solution

Mass of the car, m=1400 kgm = 1400 \text{ kg}

Initial velocity, u=54 km/h=54×518 m/s=15 m/su = 54 \text{ km/h} = 54 \times \dfrac{5}{18} \text{ m/s} = 15 \text{ m/s}

Final velocity, v=0v = 0

Height, h=10 mh = 10 \text{ m}

Initial kinetic energy, Ki=12mu2=12×1400×(15)2=157500 JK_i = \dfrac{1}{2} m u^2 = \dfrac{1}{2} \times 1400 \times (15)^2 = 157500 \text{ J}

Change in potential energy, ΔU=mgh=1400×9.8×10=137200 J\Delta U = mgh = 1400 \times 9.8 \times 10 = 137200 \text{ J}

By conservation of energy, the initial kinetic energy is used to increase the potential energy and do work against friction:

Ki=ΔU+Wagainst frictionK_i = \Delta U + W_{\text{against friction}}

157500=137200+Wagainst friction157500 = 137200 + W_{\text{against friction}}

Wagainst friction=157500−137200=20300 JW_{\text{against friction}} = 157500 - 137200 = 20300 \text{ J}
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