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If T1 and T2 are the tensions on tight and slack sides of belt in Newtons and v is the velocity in m/sec, the power transmitted is given in Watt as

  • A.\((T_{1} - T_{2})v\)
  • B.\(\frac{(T_{1} - T_{2})v}{4500}\)
  • C.\(\frac{(T_{1} - T_{2})}{60}.v\)
  • D.\(\frac{(T_{1} - T_{2})}{75}.v\)

Answer: A

The power is \((T_{1} - T_{2})v\) watts.

The net force driving the pulley is the difference of the tight-side and slack-side tensions. Power is force times velocity, so with tensions in newtons and velocity in m/s the power comes out directly in watts.

\[ P = (T_{1} - T_{2})\,v \]

Dividing by 4500 or 75 converts to horsepower in old gravitational units (kgf m/min or kgf m/s), and dividing by 60 belongs to formulas with speed in revolutions per minute. None of these is needed for watts.

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