what work must a diesel engine in an 18,000kg truck do to increase its speed from 30.m/s to 40 m/s in 5.0s? what average force did the engine exert? what power was needed (in horsepower)
work = increase in kinetic energy
= (1/2)(18,000) (40^2 -30^2)
in Joules
force * distance = work in Joules
distance = average speed * time
d = 35 * 5
so
Force = work from part a / d
power = work/time
= work from part a / 5 seconds
in Joules/sec which is Watts