Determine the average frictional force required to bring a car weighing 500 kg to a halt over a distance of 25 m , given that its initial speed is 72 km/h .

Correct answer: C. 4000 N Initial velocity u = 72 km/h = 72 × (5)/(18) m/s = 20 m/s Final velocity v = 0 Distance s = 25 m Mass m = 500 kg Using the work-energy theorem: Work done by friction = Change in kinetic energy -F × s = (1)/(2)mv^2 - (1)/(2)mu^2 -F × 25 = 0 - (1)/(2) × 500 × (20)^2 -25F = -250 × 400 25F = 100000 F = 4000 N

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

Determine the average frictional force required to bring a car...

Determine the average frictional force required to bring a car weighing

500 \text{ kg}

to a halt over a distance of

25 \text{ m}

, given that its initial speed is

72 \text{ km/h}

Options

A $8000 \text{ N}$
B $2000 \text{ N}$
C $4000 \text{ N}$
D $400 \text{ N}$

Correct answer

C. $4000 \text{ N}$

Step-by-step solution

Initial velocity

u = 72 \text{ km/h} = 72 \times \dfrac{5}{18} \text{ m/s} = 20 \text{ m/s}

Final velocity

v = 0

Distance

s = 25 \text{ m}

Mass

m = 500 \text{ kg}

Using the work-energy theorem:

Work done by friction = Change in kinetic energy

-F \times s = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2

-F \times 25 = 0 - \dfrac{1}{2} \times 500 \times (20)^2

-25F = -250 \times 400

25F = 100000

F = 4000 \text{ N}

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