Question

An object has a mass of `6 kg`. The object's kinetic energy uniformly changes from `72 KJ` to `120 KJ` over `t in [0, 3 s]`. What is the average speed of the object?

Answer 1

`v_a=8/3 m/s`

Explanation:

`Delta E_k=E_"k l"-E_"k i"" " "changing of the kinetic energy of object "`
`1/2m(v_l^2-v_i^2)=120-72`
`1/2*cancel(6)(v_l^2-v_i^2)=cancel(48)`
`(v_l^2-v_i^2)=16`
`(v_l-v_i)(v_l+v_i)=16`
`v_l-v_i=2`
`v_l+v_i=8`
`2*v_l=10" "v_l=5 m/s`
`5-v_i=2`
`v_i=3 m/2`
`a=(Delta v)/(Delta t)=(v_l-v_i)/(Delta t)=(5-3)/(3-0)=1 m/s^2" (acceleration of object)"`
`v_l^2=v_i^2+2*a*Delta x`
`5^2=3^2+2*1*Delta x`
`25=9+2*Delta x`
`16=2*Delta x`
`Delta x=8 m" displacement for (0-3)"`
`"Average velocity of an object is given by:"`
`v_a=("Total displacement")/("Total time")`
`v_a=8/3 m/s`