Power

Scientific work is done when a force exerted to an object causes it to move to a distance, \(W=F\cdot d\). We do work everytime by measuring how much force we applied in moving objects but the question is - how much work is done every time? Calculating the rate of the work done in a given time is what we call power. It is mathematically written as

\(P=\frac Wt\)                    (1)

Substituting the work as the force applied times the distance of the motion, (1) can be expressed as

\(P=\frac {F\cdot d}{t}\)                   (2).

   The equation for power illustrates that having more power means more work is done per amount of time or greater power means doing much work for a short time.

If the work is done at an angle, the power spent to an object can be calculated using the equation

\(P=\frac {F\cos \theta\cdot d }{t}\)                 (3).

If given the velocity, \(v=\frac dt\) of the object with the force applied on it, power is solved by

\(P=Fv\)                      (4).

The unit for power is \(\frac {N\cdot m}{s}=\frac Js\) or Watts, W .


Example 1.

A tow truck have to pull a 550-kg damaged car to a repair shop 1 km from the location of the car. If the truck was able to deliver the car for 50 minutes, how much work is done by the truck? How much power does the tow truck spent?

Given:

\(m_{car}=550\;kg\\ d=1\;km\\ t=50\;mins\)


Solution 1.

First, we do the conversion of units for distance and time.

\(d=1\;km \times \frac {1000\;m}{1\;km}=1000\;m\)

\(t=50\;mins\times \frac{60\;s}{1\;min}=3 \times 10^3\;s\)


Solution 2.

Solve for the weight of the car.

\(F=mg=(550\;kg)(9.8\;m/s)=5.39\times 10^3\;N\)


Solution 3.

The work done on the car is

\(W=F\cdot d\\ \quad=(5.39 \times 10^3\;N)(1000\;m)\\ \quad=5.39\times 10^6\;J.\)


Solution 4.

The power spent by the tow truck in pulling the car is

\(P=\frac Wt=\frac {5.39 \times 10^6\;J}{3.0 \times 10^3\;s}=1.80\times 10^3\;W.\)


Example 2.

The elevator is out of service thus Faber had to use the ladder in going up his condominum unit at the 3rd floor. If he weighs 637 N, how much work did he do in going up his unit taking the 3.9-m ladder inclined at 60° angle in every floor? Calculate the power expended by Faber after reaching his unit in 10 minutes.

Given:

\(F=637\;N\\ d=3.9\;m\\ \theta=60^\circ\\ t=10\;mins\)


Solution 1.

Since Faber's unit is located at the 3rd floor, the total distance covered is \(d=3.9\;m \times 2=7.8\;m\). Hence, the work done by him is

\(W=F\cos \theta \cdot d=(637\;N)(\cos 60^\circ)(7.8\;m)=2,484.3\;J.\)


Solution 2.

The time it took Faber to reach his unit is \(t=10\;mins \times \frac{60\;s}{1\;min}=600\;s.\) The power expended by him in reaching his unit is

\(P=\frac Wt=\frac{2,484.3\;J}{600\;s}=4.14\;W.\)