Question

Find the work done when the speed of a 3.00 kg object increases from rest to 23.0 m/sec.

Answer 1

793.5 Joules

Step-by-step explanation:

The work–energy theorem states that the net work done on an object equals the object’s change in kinetic energy.

So we can say W=ΔKE.

The equation for kinetic energy is

`KE=(1)/(2)mv^2`

Therefore

`W=(1)/(2)mv_f^2-(1)/(2)mv_i^2`

`W=(mv_f^2)/(2)-(mv_i^2)/(2)`

We are given

`m=3`

`v_i=0`

`v_f=23`

Lets solve for
`W`.

`W=(3*23^2)/(2)-(3*0^2)/(2)`

`W=(3*23^2)/(2)`

`W=(3*529)/(2)`

`W=(1587)/(2)`

`W=793.5`