Question

An object has a mass of `6 kg`. The object's kinetic energy uniformly changes from `18 KJ` to `4KJ` over `t in [0, 9 s]`. What is the average speed of the object?

Answer 1

The average speed is `=59.38ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The initial velocity is `=u_1`

`1/2m u_1^2=18000J`

The final velocity is `=u_2`

`1/2m u_2^2=4000J`

Therefore,

`u_1^2=2/6*18000=6000m^2s^-2`

and,

`u_2^2=2/6*4000=1333.3m^2s^-2`

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

The points are `(0,6000)` and `(9,1333.3)`

The equation of the line is

`v^2-6000=(1333.3-6000)/9t`

`v^2=-518.5t+6000`

So,

`v=sqrt((-518.5t+6000)`

We need to calculate the average value of `v` over `t in [0,9]`

`(9-0)bar v=int_0^9sqrt(-518.5t+6000) dt`

`9 barv=[(-518.5t+6000)^(3/2)/(3/2*-518.5)]_0^9`

`=((-518.5*9+6000)^(3/2)/(-777.8))-((-518.5*0+6000)^(3/2)/(-777.8))`

`=1333.5^(3/2)/-777.8-6000^(3/2)/-777.8`

`=1/778.5(6000^(3/2)-1333.5^(3/2))`

`=534.4`

So,

`barv=534.4/9=59.38ms^-1`