Question

If a `2 kg` object is constantly accelerated from `0m/s` to `16 m/s` over 6 s, how much power musy be applied at `t=3`?

Answer 1

Instantaneous power `P=42.666W`atts.

Explanation:

When a body of mass`'m'` is initially at rest, is moved by the application of a constant force, its velocity changes to `'v_o'` in time `'t_o'`,then Instantaneous power at any time `'t'` generated is given as `P=m(v_o/t_o)^2t`
Substituting the given values we get,
`P=2kg[(16m//s)/(6s)]^2xx3s=2kg[64m^2//9s^3]xx3s=(128kgm^2)/(3s^3)=42.666W`
`:.P`ower`=42.6W`atts.

Answer 2

THe power is `=42.67W`

Explanation:

To calculate the acceleration, apply the equation of motion

`v=u+at`

The initial velocity is `u=0ms^-1`

The final velocity is `v=16ms^-1`

The time is `t=6s`

The acceleration is

`a=(v-u)/t=(16-0)/6=8/3ms^-2`

The velocity at `t=3s` is

`v(3)=u+at`

`v(3)=0+8/3*3=8ms^-1`

The mass is `m=2kg`

According to Newton's Second Law

`F=ma`

The force is `F=2*8/3=16/3N`

The power is

`P=Fv=16/3*8=42.67W`

Answer 3

43 W at t=3s or
`22` W average power in the first 3 s.

Explanation:

`barP= "Average Power" = "Work"/"Time duration" =W/(Deltat)`.

`because W= FDeltax` if F is a constant.

`bar P = W/(Deltat) = F(Deltax)/(Deltat) = F barv`

The instantaneous power at a given time is then

`P = lim_(Deltat rarr0) F(Deltax)/(Deltat)=F v`

Since the object is accelerating

`F =F_"net"= ma = m (v-v_0)/t = 2((16-0)/6) = 5.33N`
`v_0=0 rArr`
`v= v_0+ at =at`
and
`P = Fv = (ma)at = ma^2t`

at t = 3s

`P = ma^2t= 2kg*((16m/s-0)/(6s))^2*3s =42.7 (kgm^2/s^2)/s ~~ 43 "Watts"`

Bear in mind that this is instantaneous power. If the question meant for average power delivered from t=0 to t=3 s, then

`barP ~~½ (43) ~~ 22 W`

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