An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 144 KJ to 120KJ over t in [0, 2 s]. What is the average speed of the object?

1 Answer
Aug 14, 2017

The average speed is =256.8ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is =4kg

The initial velocity is =u_1ms^-1

The final velocity is =u_2 ms^-1

The initial kinetic energy is 1/2m u_1^2=144000J

The final kinetic energy is 1/2m u_2^2=120000J

Therefore,

u_1^2=2/4*144000=72000m^2s^-2

and,

u_2^2=2/4*120000=60000m^2s^-2

The graph of v^2=f(t) is a straight line

The points are (0,72000) and (2,60000)

The equation of the line is

v^2-72000=(60000-72000)/2t

v^2=-6000t+72000

So,

v=sqrt((-6000t+72000)

We need to calculate the average value of v over t in [0,2]

(2-0)bar v=int_0^2(sqrt(-6000t+72000))dt

2 barv=[((-6000t+72000)^(3/2)/(3/2*-6000))]_0^2

=((-6000*2+72000)^(3/2)/(-9000))-((-6000*0+72000)^(3/2)/(-9000))

=72000^(3/2)/9000-60000^(3/2)/9000

=513.6

So,

barv=513.6/2=256.8ms^-1

The average speed is =256.8ms^-1