Question

An object has a mass of `3 kg`. The object's kinetic energy uniformly changes from `75 KJ` to `42 KJ` over `t in [0, 15 s]`. What is the average speed of the object?

Answer 1

The average speed is `=181.3ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The initial velocity is `=u_1`

`1/2m u_1^2=75000J`

The final velocity is `=u_2`

`1/2m u_2^2=42000J`

Therefore,

`u_1^2=2/3*75000=50000m^2s^-2`

and,

`u_2^2=2/3*42000=28000m^2s^-2`

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

The points are `(0,50000)` and `(15,28000)`

The equation of the line is

`v^2-50000=(28000-50000)/15t`

`v^2=-1466.7t+50000`

So,

`v=sqrt((-1466.7t+50000)`

We need to calculate the average value of `v` over `t in [0,15]`

`(15-0)bar v=int_0^15sqrt((-1466.7t+50000)dt`

`15 barv=[((-1466.7t+50000)^(3/2)/(3/2*-1466.7)]_0^15`

`=((-1466.7*15+50000)^(3/2)/(-2200))-((-1466.7*0+50000)^(3/2)/(-2200))`

`=30000^(3/2)/-2200-50000^(3/2)/-2200`

`=50000^(3/2)/2200-30000^(3/2)/2200`

`=2720.1`

So,

`barv=2720.1/15=181.3ms^-1`