Question

An object has a mass of `6 kg`. The object's kinetic energy uniformly changes from `225 KJ` to `150 KJ` over `t in [0, 8 s]`. What is the average speed of the object?

Answer 1

The average speed is `=249.6ms^-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=225000J`

The final kinetic energy is `1/2m u_2^2=150000J`

Therefore,

`u_1^2=2/6*225000=75000m^2s^-2`

and,

`u_2^2=2/6*150000=50000m^2s^-2`

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

The points are `(0,75000)` and `(8,50000)`

The equation of the line is

`v^2-75000=(50000-75000)/8t`

`v^2=-3125t+75000`

So,

`v=sqrt(-3125t+75000)`

We need to calculate the average value of `v` over `t in [0,8]`

`(8-0)bar v=int_0^8(sqrt(-3125t+75000))dt`

`8 barv= [(-3125t+75000)^(3/2)/(-3/2*3125)] _( 0) ^ (8)`

`=((-3125*8+75000)^(3/2)/(-4687.5))-((-3125*0+75000)^(3/2)/(-4687.5))`

`=75000^(3/2)/4687.5-50000^(3/2)/4687.5`

`=1996.6`

So,

`barv=1996.6/8=249.6ms^-1`

The average speed is `=249.6ms^-1`

Answer 2

`248.73 m/s`

Explanation:

`KE=1/2mv^2`

`v=sqrt((2KE)/m)`

So, v initial `=(sqrt((2*225,000J)/(6 kg))=273.86 m/s`

and v final`=(sqrt((2*150,000J)/(6 kg))=223.607 m/s`

`v avg=((v f+v i)/2)=((273.86 + 223.607 m/s)/2)=248.73 m/s`