1 A 4000 lbm vehicle accelerates from rest to 60 mph During

1. A 4,000 lbm vehicle accelerates from rest to 60 mph During the acceleration, the vehicle climbs 10 feet in elevation. Determine a. The change in kinetic energy of the vehicle (BTU) b. The change in potential energy of the vehicle(BTU) c. The required work(BTU) d. The required power (hp) to achieve this in 7 seconds.

Solution

a)

1 lbm =0.453592 kg

m=4000*0.453592 =1814.4 kg

final speed ,V_f=60 mph

1 mph =0.44704 m/s

V_f=60*0.44704 =26.8 m/s

SInce it starts from rest V_i=0

Change in Kinetic energy

KE =(1/2)m(V_f²-V_i²)

KE=(1/2)*1814.4*26.8² =652.7 KJ

1 KJ =0.947817

KE =652.7*0.947817

KE=618.6 BTU

b)

1 feet =0.3048 m

h=10*0.3048 =3.048 m

Change in potential energy of the vehicle

PE =mgh =1814.4*9.81*3.048

PE=54.25 KJ

In BTU

PE =54.25*0.947817 =51.4 BTU

c)

Required work done is

W =KE+PE =670 BTU

d)

Required power is

P=W/t =670/7 =95.7 BTU/s

1 BTU/s =1.4148532 hp

P=95.7*1.4148532

P=135.4 hp