A scooter, which together with its rider weighs 150 kg , is travelling at 36 km/hr . Determine the horizontal force on the scooter needed to successfully navigate a turn of radius 30 m .

Correct answer: A. 500 N Mass of the scooter and rider, m = 150 kg Velocity of the scooter, v = 36 km/hr = 36 × (5)/(18) m/s = 10 m/s Radius of the turn, r = 30 m The horizontal force required to navigate the turn is the centripetal force, which is given by: F = (mv^2)/(r) Substituting the given values: F = (150 × (10)^2)/(30) F = (150 × 100)/(30) = 500 N

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Circular Motion

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Circular Motion

A scooter, which together with its rider weighs 150 kg , is...

A scooter, which together with its rider weighs

150 \text{ kg}

, is travelling at

36 \text{ km/hr}

. Determine the horizontal force on the scooter needed to successfully navigate a turn of radius

30 \text{ m}

Options

A $500 \text{ N}$
B $50 \text{ N}$
C $6480 \text{ N}$
D $180 \text{ N}$

Correct answer

A. $500 \text{ N}$

Step-by-step solution

Mass of the scooter and rider,

m = 150 \text{ kg}

Velocity of the scooter,

v = 36 \text{ km/hr} = 36 \times \dfrac{5}{18} \text{ m/s} = 10 \text{ m/s}

Radius of the turn,

r = 30 \text{ m}

The horizontal force required to navigate the turn is the centripetal force, which is given by:

F = \dfrac{mv^2}{r}

Substituting the given values:

F = \dfrac{150 \times (10)^2}{30}

F = \dfrac{150 \times 100}{30} = 500 \text{ N}

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