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An air compressor is powered by a $200 \mathrm{rad} / \mathrm{s}$ electric motor using a V-belt drive. The motor pulley is $8 \mathrm{~cm}$ in radius, and the tension in the $\mathrm{V}$-belt is $135 \mathrm{~N}$ on one side and $45 \mathrm{~N}$ on the other. The power of the motor will be
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Verified Answer
The correct answer is:
$1.44 \mathrm{~kW}$
Since the net force on the belt is the difference between the two tensions, the torque exerted by the motor on the belt is,
$$
\begin{aligned}
\tau & =F r=(135 \mathrm{~N}-45 \mathrm{~N})(0.08 \mathrm{~m}) \\
& =7.2 \mathrm{~N} \mathrm{~m}
\end{aligned}
$$
The power output of the motor is therefore
$$
\begin{aligned}
P & =\tau \omega=(72 \mathrm{~N} \cdot \mathrm{m})\left(200 \mathrm{rad} \mathrm{s}^{-1}\right) \\
& =1440 \mathrm{~W}=1.44 \mathrm{~kW}
\end{aligned}
$$
$$
\begin{aligned}
\tau & =F r=(135 \mathrm{~N}-45 \mathrm{~N})(0.08 \mathrm{~m}) \\
& =7.2 \mathrm{~N} \mathrm{~m}
\end{aligned}
$$
The power output of the motor is therefore
$$
\begin{aligned}
P & =\tau \omega=(72 \mathrm{~N} \cdot \mathrm{m})\left(200 \mathrm{rad} \mathrm{s}^{-1}\right) \\
& =1440 \mathrm{~W}=1.44 \mathrm{~kW}
\end{aligned}
$$
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