Question

An object has a mass of `6 kg`. The object's kinetic energy uniformly changes from `84 KJ` to `48 KJ` over `t in [0, 6 s]`. What is the average speed of the object?

Answer 1

The average speed is `=147.9ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The mass is `m=6kg`

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=84000J`

The final kinetic energy is `1/2m u_2^2=48000J`

Therefore,

`u_1^2=2/6*84000=28000m^2s^-2`

and,

`u_2^2=2/6*48000=16000m^2s^-2`

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

The points are `(0,28000)` and `(6,16000)`

The equation of the line is

`v^2-28000=(16000-28000)/6t`

`v^2=-2000t+28000`

So,

`v=sqrt(-2000t+28000)`

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

`(6-0)bar v=int_0^6(sqrt(-2000t+28000))dt`

`6 barv= [(-2000t+28000)^(3/2)/(-3/2*2000)] _( 0) ^ (6)`

`=((-2000*6+28000)^(3/2)/(-3000))-((-2000*0+28000)^(3/2)/(-3000))`

`=28000^(3/2)/3000-16000^(3/2)/3000`

`=887.15`

So,

`barv=887.15/6=147.9ms^-1`

The average speed is `=147.9ms^-1`