# 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 ?

Apr 6, 2016

$P \left(1\right) = \frac{16}{49} \frac{J}{s} = \frac{16}{49} \text{Watts}$

#### Explanation:

Given: Object mass, $m = 12 k g$, ${v}_{i} = 0 , {v}_{f} = 4 \frac{m}{s}$

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

Solution Strategy:
a) Kinematic equation to determine the acceleration: ${v}_{f} = {v}_{i} + a t$
b) Find the kinetic energy at any timer $\frac{1}{2} {v}^{2} \left(t\right)$
c) Calculate Power using: $P \left(t\right) = \frac{\mathrm{dW}}{\mathrm{dt}} {|}_{t = 1} = P \left(1\right)$

$\textcolor{b r o w n}{\text{a)}}$
v(t) = at; v(7)=v_f=4=7a
$a = \frac{4}{7} \frac{m}{s} ^ 2$ thus
$v \left(t\right) = \frac{4}{7} t$

$\textcolor{b r o w n}{\text{b)}}$ Kinetic energy = Work
$W = \frac{1}{m} {v}^{2} \left(t\right) = \frac{1}{2} \cdot {\left(\frac{4}{7} t\right)}^{2} = \frac{8}{49} {t}^{2}$

$\textcolor{b r o w n}{\text{c)}}$ Power, $P = \frac{\mathrm{dW}}{\mathrm{dt}}$
$P \left(t\right) = \frac{8}{49} \frac{d \left({t}^{2}\right)}{\mathrm{dt}} = \frac{16}{49} t$ Evaluate power, $P \left(t\right) {|}_{t = 1} = P \left(1\right)$
$P \left(1\right) = \frac{16}{49} \frac{J}{s} = \frac{16}{49} \text{Watts}$

Apr 12, 2016

$= 3.92 J {s}^{-} 1$, rounded to second decimal place.

#### Explanation:

For constant acceleration, the kinematic expression between velocity, acceleration and time is
$v = u + a t$, Inserting given values and solving for acceleration
$4 = 0 + a \times 7$
or $a = \frac{4}{7} m {s}^{-} 2$
We know that Force $F = m a$
or $F = 12 \times \frac{4}{7} = 6.86 N$ 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 $\mathrm{dx}$ in time $\mathrm{dt}$
Work done $\mathrm{dW} = \vec{F} \cdot \vec{\mathrm{ds}}$, if force and distance moved are in the same direction
Power applied $\frac{\mathrm{dw}}{\mathrm{dt}} = F \cdot \frac{\mathrm{ds}}{\mathrm{dt}}$
$= F \cdot a t$, $\left(\therefore \frac{\mathrm{ds}}{\mathrm{dt}} = v\right)$
Power applied at $t = 1$
$= F \cdot a t {|}_{t = 1}$

$P {|}_{t = 1} = 6.86 \times \frac{4}{7} \times 1$
$= 3.92 J {s}^{-} 1$ rounded to second decimal place.