Power in physics

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The quantity work has to do with a force causing a displacement in the energy of a system. The equation for work is as follows:

W = Fd

[edit] Fd

The term Fd in itself is actually used to describe something done to the object. Something in the environment put forth a force, F, which displaced the object a distance d. The work-energy theorem states that when work is done on an object, it results in a change in kinetic energy. This relationship was established by physicist James Prescott Joule. However, when discussing work it should be mentioned that work has nothing to do with the amount of time that this force acts to cause the displacement. Sometimes, the work is done very quickly and other times the work is done rather slowly. For example, a rock climber takes an abnormally long time to elevate her body up a few meters along the side of a cliff. On the other hand, a trail hiker (who selects the easier path up the mountain) might elevate her body a few meters in a short amount of time. The two people might do the same amount of work, yet the hiker does the work in considerably less time than the rock climber. The quantity which has to do with the rate at which a certain amount of work is done is known as the power. The hiker has a greater power rating than the rock climber.

[edit] Work/Time Ratio

Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is computed using the following equation.

Power = Work / time

[edit] Watt

The standard metric unit of power is the Watt. As is implied by the equation for power, a unit of power is equivalent to a unit of work divided by a unit of time. Thus, a Watt is equivalent to a Joule/second. For historical reasons, the horsepower is occasionally used to describe the power delivered by a machine. One horsepower is equivalent to approximately 750 Watts. Because a watt is such a small unit, we often measure power in kilowatts (kW) instead. This is equivalent to 1000 Watts.

[edit] Ben Pumpiniron elevates

Suppose a guy named Ben Pumpiniron elevates his 80-kg body up the 2.0 meter stairwell in 1.8 seconds. If this were the case, then we could calculate Ben's power rating. It can be assumed that Ben must apply a 800-Newton downward force upon the stairs to elevate his body. By so doing, the stairs would push upward on Ben's body with just enough force to lift his body up the stairs. It can also be assumed that the angle between the force of the stairs on Ben and Ben's displacement is 0 degrees. With these two approximations, Ben's power rating could be determined as shown below.

Power = Work /time = (800 N * 2.0 m)/ 1.8 secs

Power = 889 Watts

Ben's power rating is 889 Watts; what a "horse."

[edit] Force and displacement

The expression for power is work/time. Now since the expression for work is force*displacement, the expression for power can be rewritten as (force*displacement)/time. Yet since the expression for velocity is displacement/time, the expression for power can be rewritten once more as force*velocity. This is shown below.

Power = Work / time = Force * (displacement / time)

Power = Force * (displacement / time)

Power = Force * Velocity

[edit] Strong and fast

This new expression for power reveals that a powerful machine is both strong (big force) and fast (big velocity). The powerful car engine is strong and fast. The powerful farm equipment is strong and fast. The powerful weightlifters are strong and fast. The powerful linemen on a football team are strong and fast. A machine which is strong enough to apply a big force to cause a displacement in a small mount of time (i.e., a big velocity) is a powerful machine.

[edit] Sample Power and Work Word Problems

Use your understanding of work and power to answer the following questions.

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. During the Personal Power lab, Jack and Jill ran up the hill. Jack is twice as massive as Jill; yet Jill ascended the same distance in half the time. Who did the most work? Who delivered the most power? Explain your answers.
3. A tired squirrel (mass of 1 kg) does push-ups by applying a force to elevate its center-of-mass by 5 cm. Determine the number of push-ups which a tired squirrel must do in order to do a mere 1.0 Joule of work. If the tired squirrel does all this work in 4 seconds, then determine its power.
4. If little Nellie Newton lifts her 40-kg body a distance of 0.25 meters in 2 seconds, then what is the power delivered by little Nellie's biceps?
5. 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-meters above the first floor. The average passenger's mass is 60 kg. Determine the power requirement of the escalator in order to move this number of passengers in this amount of time.
6. During a game of golf, the player exerts a 5.5-N force over a distance of .2m. How much work has the golf player done on the ball? What is the change in the energy of the ball?

[edit] Answers

1. Ben and Will do the same amt. of work; they use the same force to lift the same barbell the same distance above their heads. Yet, Ben is the most "powerful" since he does the same work in less time. Power and time are inversely proportional.
2. Jack does more work than Jill. Jack must apply twice to force to lift his twice-as-massive body up the same set of stairs. Yet, Jill is just as "powerful" as Jack. Jill does one-half the work yet does it in one-half the time. The reduction of work done is compensated for the reduction in time.
3. The tired squirrel does 1 Joule of work in 4.0 sec. The power rating of this squirrel is found by P = W/t = (1J/ 4.0sec) = 0.25 Watts.
4. The work done to lift her body is W = F * d = 400 N * 0.25 m , W = 100 J. The power is the work/time ratio which is 100 J/ 2 seconds = 50 Watts.
5. The work done to lift one passenger is 600 N * 5 m = 3000J. The work done to lift twenty passengers is 20 * 3000 J = 60,000 Joules. The power is the work/time ratio which is 60,000 J / 60 seconds = 1000 Watts.
6. The work done on the ball is: W=Fd = (5.5-N) * (.2m) = 1.1J. This is the work. Basically we are also trying to find the change in K of the ball, so refer back to the work-energy theorem. Change in K = W = 1.1J.

[edit] Useful Site/Book

• Physics Classroom
• Zitzewitz, Paul W.. Physics: Principles and Problems. Ohio: Glencoe/McGraw-Hill, 1999.