An object has a mass of 5 kg. The object's kinetic energy uniformly changes from 55 KJ to 42KJ over t in [0,4s]. What is the average speed of the object?

1 Answer
Narad T.
Nov 30, 2017

The average speed is =139.2ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is m=5kg

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=55000J

The final kinetic energy is 1/2m u_2^2=42000J

Therefore,

u_1^2=2/5*55000=22000m^2s^-2

and,

u_2^2=2/5*42000=16800m^2s^-2

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

The points are (0,22000) and (4,16800)

The equation of the line is

v^2-22000=(16800-22000)/4t

v^2=-1300t+22000

So,

v=sqrt(-1300t+22000)

We need to calculate the average value of v over t in [0,4]

(4-0)bar v=int_0^4(sqrt(-1300t+22000))dt

4 barv=[((-1300t+22000)^(3/2)/(-3/2*1300))]_0^4

=((-1300*4+22000)^(3/2)/(-1950))-((-1300*0+22000)^(3/2)/(-1950))

=22000^(3/2)/1950-16800^(3/2)/1950

=556.7

So,

barv=556.7/4=139.2ms^-1

The average speed is =139.2ms^-1

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