Question

A 3000 kg car goes from 0 to 70 miles per hour in 6 seconds. Assume all the work supplied by the engine goes into increasing the kinetic energy of the car.

1. If the engine is purely electrical (and 100% efficient), determine the electrical work supplied in kW.h by the engine battery. A kW.h of electricity costs about 25 cents (this is just for your information).

(kWh)

2.If the engine is a gasoline powered engine, determine the fuel consumed in kg if the engine has an efficiency of 40% and the heating value of gasoline is 45 MJ/kg.

(kg)

Answer 1

Mass of car M=3000 kg

Initial velocity of car, v_i=0

Final velocity of car, v_f=70 mile/h=(70$\times$(1.61$\times$1000))$\div$(3600)=31.2928 m/s

accorinding to work energy theorm

Work,W=change in K.E=(1/2)mv_f²-(1/2)mv_i²​=(1/2)(3000)(31.2928)²​=1468858.998 Ws

Part 1.

Electrical work in kWh is

W=1468858.998$\times$10^-3$\times$(3600)^-1=0.408 kWh

cost of 1kWh=25 cents

so cost of 0.408 kWh is=0.408$\times$25=10.20 cents

Part b)

workdone by engine=(40/100)(1468858.998)=587543.6 J

As

1 kg fuel consumed for=45MJ=45$\times$10⁶ J

so for 587543.6 j =587543.6/(45$\times$10⁶)=0.0131 kg fuel