If a $12 k g$ $12 kg$ object is constantly accelerated from $0 \frac{m}{s}$ $0m/s$ to $40 \frac{m}{s}$ $40 m/s$ over 8 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 $40 \frac{m}{s}$ $40 m/s$ over 8 s, how much power musy be applied at $t = 1$ $t=1$ ?

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Answer 1 · 2017-03-30T16:12:55

$300 W$ $300"W"$ (Watts)

Explanation:

The power applied to an object is defined as $P = \frac{W}{t}$ $P=W/t$ , and expanding the definition of work, $P = \frac{F d}{t}$ $P=(Fd)/t$ (assuming force and displacement are in the same direction). $\frac{d}{t} = v$ $d/t=v$ , so $P = F v$ $P=Fv$ . $F = m a$ $F=ma$ , so $P = m a v$ $P=mav$ .

The acceleration of an object is $(vec(v_f)-vec(v_i))/(Deltat)$ , for this object it is $(40"m/s"-0"m/s")/(8"s")=5"m/s"^2$ .

The velocity of an object at time $t$ is $at$ , so in this case the velocity at $1"s"$ is $5"m/s"^2*1"s"=5"m/s"$

Finally, solving for power using $P=mav$ , we get $P=12"kg"*5"m/s"^2*5"m/s"=300"W"$

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If a $12 k g$ $12 kg$ object is constantly accelerated from $0 \frac{m}{s}$ $0m/s$ to $40 \frac{m}{s}$ $40 m/s$ over 8 s, how much power musy be applied at $t = 1$ $t=1$ ?

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Explanation:

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If a 12 kg12kg12 kg object is constantly accelerated from 0m/s0ms0m/s to 40 m/s40ms40 m/s over 8 s, how much power musy be applied at t=1 t=1t=1 ?

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If a $12 k g$ $12 kg$ object is constantly accelerated from $0 \frac{m}{s}$ $0m/s$ to $40 \frac{m}{s}$ $40 m/s$ over 8 s, how much power musy be applied at $t = 1$ $t=1$ ?