PrepRung
Sign inSign up free

JEE Main 2016 · Physics · Work, Energy and Power · medium

A car of mass 1000 kg has an engine delivering a constant power of 40 kW on a level road. The resistive force is proportional to the speed, and the…

Single correct option

A car of mass 1000 kg1000\ \mathrm{kg} has an engine delivering a constant power of 40 kW40\ \mathrm{kW} on a level road. The resistive force is proportional to the speed, and the maximum speed of the car is 40 m s−140\ \mathrm{m\,s^{-1}}. The acceleration of the car when its speed is 20 m s−120\ \mathrm{m\,s^{-1}} is
  1. A. 2.0 m s−22.0\ \mathrm{m\,s^{-2}}
  2. B. 0.75 m s−20.75\ \mathrm{m\,s^{-2}}
  3. C. 1.5 m s−21.5\ \mathrm{m\,s^{-2}}
  4. D. 1.0 m s−21.0\ \mathrm{m\,s^{-2}}
Show answer and solution

Answer: (C) 1.5 m s−21.5\ \mathrm{m\,s^{-2}}

Let the resistive force be kvkv. At maximum speed the driving force equals the resistance: Pvmax=kvmax\dfrac{P}{v_{max}}=kv_{max}, so k=Pvmax2=400001600=25 N s m−1k=\dfrac{P}{v_{max}^2}=\dfrac{40000}{1600}=25\ \mathrm{N\,s\,m^{-1}}. At v=20 m s−1v=20\ \mathrm{m\,s^{-1}} the driving force is Pv=2000 N\dfrac Pv=2000\ \mathrm{N} and the resistance is 25×20=500 N25\times20=500\ \mathrm{N}. Newton's second law gives a=2000−5001000=1.5 m s−2a=\dfrac{2000-500}{1000}=1.5\ \mathrm{m\,s^{-2}}. Ignoring the speed dependence of the resistance, e.g. using 1000 N1000\ \mathrm{N} at maximum speed, would give a=1.0 m s−2a=1.0\ \mathrm{m\,s^{-2}}.
Report a mistake

Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake