An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 144 KJ to 120KJ over t in [0, 6 s]. What is the average speed of the object?

1 Answer
Jul 16, 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 (6,60000)

The equation of the line is

v^2-72000=(60000-72000)/6t

v^2=-2000t+72000

So,

v=sqrt((-2000t+72000)

We need to calculate the average value of v over t in [0,6]

(6-0)bar v=int_0^6sqrt(-2000t+72000))dt

6 barv=[((-2000t+72000)^(3/2)/(-3/2*2000)]_0^6

=((-2000*6+72000)^(3/2)/(-3000))-((-2000*0+72000)^(3/2)/(-3000))

=72000^(3/2)/3000-60000^(3/2)/3000

=1540.9

So,

barv=1540.9/6=256.8ms^-1

The average speed is =256.8ms^-1