Question

If a `12 kg` object is constantly accelerated from `0m/s` to `4 m/s` over 7 s, how much power musy be applied at `t=1`?

Answer 1

`P(1) = 16/49J/s= 16/49 "Watts"`

Explanation:

Given: Object mass, `m= 12kg`, `v_i=0, v_f=4m/s`

Required: Power applied or use up at `t=1`

Solution Strategy:
a) Kinematic equation to determine the acceleration: `v_f= v_i +at`
b) Find the kinetic energy at any timer `1/2v^2(t)`
c) Calculate Power using: `P(t)= (dW)/dt|_(t=1)=P(1)`

`color(brown)("a)")`
`v(t) = at; v(7)=v_f=4=7a`
`a=4/7 m/s^2` thus
`v(t) = 4/7t`

`color(brown)("b)")` Kinetic energy = Work
`W=1/mv^2(t)= 1/2*(4/7t)^2=8/49t^2`

`color(brown)("c)")` Power, `P= (dW)/dt`
`P(t)= 8/49 (d(t^2))/dt= 16/49t` Evaluate power, `P(t)|_(t=1)=P(1)`
`P(1) = 16/49J/s= 16/49 "Watts"`

Answer 2

`=3.92Js^-1`, rounded to second decimal place.

Explanation:

For constant acceleration, the kinematic expression between velocity, acceleration and time is
`v=u+at`, Inserting given values and solving for acceleration
`4=0+axx7`
or `a=4/7ms^-2`
We know that Force `F=ma`
or `F=12xx4/7=6.86N` rounded to two decimal places.
It is given that during the period acceleration is constant.
Implies that force is also constant during the period.

Let the object move through a distance `dx` in time `dt`
Work done `dW=vecF cdotvecds`, if force and distance moved are in the same direction
Power applied `(dw)/(dt)=Fcdot(ds)/dt`
`=Fcdot at`, `( :. (ds)/dt=v)`
Power applied at `t=1`
`=Fcdot at|_(t=1)`

`P|_(t=1)=6.86xx4/7xx1`
`=3.92Js^-1` rounded to second decimal place.