If a $12 k g$ $12 kg$ object is constantly accelerated from $0 \frac{m}{s}$ $0m/s$ to $4 \frac{m}{s}$ $4 m/s$ over 7 s, how much power musy be applied at $t = 1$ $t=1$ ?

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If a $12 k g$ $12 kg$ object is constantly accelerated from $0 \frac{m}{s}$ $0m/s$ to $4 \frac{m}{s}$ $4 m/s$ over 7 s, how much power musy be applied at $t = 1$ $t=1$ ?

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Answer 1 · 2016-04-06T23:32:38

$P (1) = \frac{16}{49} \frac{J}{s} = \frac{16}{49} Watts$ $P(1) = 16/49J/s= 16/49 "Watts"$

Explanation:

Given: Object mass, $m = 12 k g$ $m= 12kg$ , $v_{i} = 0, v_{f} = 4 \frac{m}{s}$ $v_i=0, v_f=4m/s$

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

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

$a)$ $color(brown)("a)")$
$v (t) = a t; v (7) = v_{f} = 4 = 7 a$ $v(t) = at; v(7)=v_f=4=7a$
$a = \frac{4}{7} \frac{m}{s^{2}}$ $a=4/7 m/s^2$ thus
$v (t) = \frac{4}{7} t$ $v(t) = 4/7t$

$b)$ $color(brown)("b)")$ Kinetic energy = Work
$W = \frac{1}{m} v^{2} (t) = \frac{1}{2} \cdot {(\frac{4}{7} t)}^{2} = \frac{8}{49} t^{2}$ $W=1/mv^2(t)= 1/2*(4/7t)^2=8/49t^2$

$c)$ $color(brown)("c)")$ Power, $P = \frac{d W}{d t}$ $P= (dW)/dt$
$P (t) = \frac{8}{49} \frac{d (t^{2})}{d t} = \frac{16}{49} t$ $P(t)= 8/49 (d(t^2))/dt= 16/49t$ Evaluate power, $P (t) ∣_{t = 1} = P (1)$ $P(t)|_(t=1)=P(1)$
$P (1) = \frac{16}{49} \frac{J}{s} = \frac{16}{49} Watts$ $P(1) = 16/49J/s= 16/49 "Watts"$

Answer 2 · 2016-04-12T07:48:58

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

Explanation:

For constant acceleration, the kinematic expression between velocity, acceleration and time is
$v = u + a t$ $v=u+at$ , Inserting given values and solving for acceleration
$4 = 0 + a \times 7$ $4=0+axx7$
or $a = \frac{4}{7} m s^{- 2}$ $a=4/7ms^-2$
We know that Force $F = m a$ $F=ma$
or $F = 12 \times \frac{4}{7} = 6.86 N$ $F=12xx4/7=6.86N$ 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 $d x$ $dx$ in time $d t$ $dt$
Work done $d W = \to F \cdot \to d s$ $dW=vecF cdotvecds$ , if force and distance moved are in the same direction
Power applied $\frac{d w}{d t} = F \cdot \frac{d s}{d t}$ $(dw)/(dt)=Fcdot(ds)/dt$
$= F \cdot a t$ $=Fcdot at$ , $( :. (ds)/dt=v)$
Power applied at $t=1$
$=Fcdot at|_(t=1)$

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

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If a $12 k g$ $12 kg$ object is constantly accelerated from $0 \frac{m}{s}$ $0m/s$ to $4 \frac{m}{s}$ $4 m/s$ over 7 s, how much power musy be applied at $t = 1$ $t=1$ ?

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If a 12 kg12kg12 kg object is constantly accelerated from 0m/s0ms0m/s to 4 m/s4ms4 m/s over 7 s, how much power musy be applied at t=1 t=1t=1 ?

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Explanation:

Explanation:

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If a $12 k g$ $12 kg$ object is constantly accelerated from $0 \frac{m}{s}$ $0m/s$ to $4 \frac{m}{s}$ $4 m/s$ over 7 s, how much power musy be applied at $t = 1$ $t=1$ ?