Work and Power By: Jeffrey Chen and Daniel Lee

What is Work? In physics, mechanical work is a scalar quantity that can be described as the product of a force times the distance through which it acts, and it is called the work of the force. Only the component of a force in the direction of the movement of its point of application does work. In other words, work is when a force acts upon an object to cause a displacement of the object

How to calculate work? Mathematically, work can be expressed by the following equation.where F is the force, d is the displacement, and the angle (theta) is defined as the angle between the force and the displacement vector. W = F x d x Cos(Theta) = change in total energy

What is Power? In physics, power is the rate at which energy is transferred, used, or transformed. For example, the rate at which a light bulb transforms electrical energy into heat and light is measured in watts—the more wattage, the more power, or equivalently the more electrical energy is used per unit time. Energy transfer can be used to do work, so power is also the rate at which this work is performed. The output power of an electric motor is the product of the torque the motor generates and the angular velocity of its output shaft.

How to calculate power? In order to calculate the Power we have the formula given by, P = W / t or Power = Work or Energy in Joule divided by Time W = Energy with units of Joules (J) time = t with units of sec The units of Power is J/s that is Watt (W). Another formula which can be used to calculate the power is, P = I^2 *R where, I is the current in Amperes or its sub-units and R is the resistance in Ohms Power can be calculated by, P = V I where, V is the Voltage in Volts and I is the current in Amperes. Furthermore, P = V^2 /R where, V is the Voltage in Volts and R is the Resistancel.

Work Problems

Answers! Diagram A Answer: W = (100 N) * (5 m)* cos(0 degrees) = 500 J The force and the displacement are given in the problem statement. It is said (or shown or implied) that the force and the displacement are both rightward. Since F and d are in the same direction,the angle is 0 degrees. Diagram B Answer: W = (100 N) * (5 m) * cos(30 degrees) = 433 J The force and the displacement are given in theproblem statement. It is said that the displacement is rightward. It is shown that the force is 30 degrees above the horizontal. Thus, the angle between F and d is 30 degrees. Diagram C Answer: W = (147 N) * (5 m) * cos(0 degrees) = 735 J The displacement is given in the problem statement. The applied force must be 147 N since the 15-kg mass (F grav =147 N) is lifted at constant speed. Since F and d are in the same direction, the angle is 0 degrees.

Power Problems 1. Two physics students, Will N. Andable and Ben Pumpiniron, are in the weightlifting room. Will lifts the 100-pound barbell over his head 10 times in one minute; Ben lifts the 100-pound barbell over his head 10 times in 10 seconds. Which student does the most work? ______________ Which student delivers the most power? ______________ Explain your answers. 2. A tired squirrel (mass of approximately 1 kg) does push-ups by applying a force to elevate its center-of-mass by 5 cm in order to do a mere 0.50 Joule of work. If the tired squirrel does all this work in 2 seconds, then determine its power. 3. An escalator is used to move 20 passengers every minute from the first floor of a department store to the second. The second floor is located 5.20 meters above the first floor. The average passenger's mass is 54.9 kg. Determine the power requirement of the escalator in order to move this number of passengers in this amount of time.

Power Answers 1. Ben and Will do the same amount of work. They apply the same force to lift the same barbell the same distance above their heads. Yet, Ben is the most "power-full" since he does the same work in less time. Power and time are inversely proportional. 2. The tired squirrel does 0.50 Joule of work in 2.0 seconds. The power rating of this squirrel is found by P = W / t = (0.50 J) / (2.0 s) = 0.25 Watts 3. A good strategy would involve determining the work required to elevate one average passenger. Then multiply this value by 20 to determine the total work for elevating 20 passengers. Finally, the power can be determined by dividing this total work value by the time required to do the work. The solution goes as follows: W 1 passenger = F d cos(0 deg) W 1 passenger = (54.9 kg 9.8 m/s 2 ) 5.20 m = 2798 J (rounded) W 20 passengers = J (rounded) P = W 20 passengers / time = (55954 J) / (60 s) P = 933 W