Question

An object has a mass of `4 kg`. The object's kinetic energy uniformly changes from `144 KJ` to `120KJ` over `t in [0, 2 s]`. What is the average speed of the object?

Answer 1

The average speed is `=256.8ms^-1`

Explanation:

The kinetic energy is

`KE=1/2mv^2`

The mass is `=4kg`

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=144000J`

The final kinetic energy is `1/2m u_2^2=120000J`

Therefore,

`u_1^2=2/4*144000=72000m^2s^-2`

and,

`u_2^2=2/4*120000=60000m^2s^-2`

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

The points are `(0,72000)` and `(2,60000)`

The equation of the line is

`v^2-72000=(60000-72000)/2t`

`v^2=-6000t+72000`

So,

`v=sqrt((-6000t+72000)`

We need to calculate the average value of `v` over `t in [0,2]`

`(2-0)bar v=int_0^2(sqrt(-6000t+72000))dt`

`2 barv=[((-6000t+72000)^(3/2)/(3/2*-6000))]_0^2`

`=((-6000*2+72000)^(3/2)/(-9000))-((-6000*0+72000)^(3/2)/(-9000))`

`=72000^(3/2)/9000-60000^(3/2)/9000`

`=513.6`

So,

`barv=513.6/2=256.8ms^-1`

The average speed is `=256.8ms^-1`