PHY11L - E201 WORK, ENERGY, AND POWER

EXPERIMENT 201:  WORK, ENERGY, AND POWER

 

 

 

 DATA and OBSERVATIONS

TABLE 1.   Part 1. Determining the Force, Work and Power of the Fan cart.
Force of the Fan Cart = weight of pan +weight added  =        0.294      Newtons
Trial	Displacement,	Time,	Work	Power
1	0.40              m	0.64            sec	0.12          Joules	0.18          Watts
2	0.45              m	0.68            sec	0.13          Joules	0.19           Watts
3	0.50              m	0.74            sec	0.15          Joules	0.20          Watts
4	0.55              m	0.79            sec	0.16          Joules	0.20          Watts

 

 

TABLE 2. Part 2. Work by a Force on a Curved Path.
Length of string,  =     0.325      m			Weight of mass,   =      4.90    N			Initial height,  =   0.070  m
Trial	Force	Final height,	Increase in height,	Angle,	Displacement,	Work	Gravitational potential energy
1	2.5      N	0.10   m	0.30   m	30	0.16    m	0.21    Joules	0.49    Joules
2	5.4      N	0.15   m	0.80   m	50	0.25    m	0.57    Joules	0.74    Joules
3	7.4      N	0.24   m	0.17     m	70	0.31     m	1.05     Joules	1.18     Joules
4	13.0      N	0.38   m	0.31     m	90	0.33    m	1.60     Joules	1.86    Joules

 

 

 

                                                                                    Approved by:

 

instructor	date

 

 

 

  GUIDE QUESTIONS

 

          Part 1.

 

1.    In part 1, should the work done be increasing every trial? Explain.

 

No, the work done should be decreasing every trial. Since work is directly proportional to the force applied by the fan based from the data on the data sheet, the work is also directly proportional to the displacement of the fan cart.  Since our group decreases the displacement by 10 cm for every trial while the cart is at constant velocity, the time elapsed also decreases. The work done is given by the formula;

 

W = Fs              where s = displacement

                      F = Force

 

 

Since the force in this experiment is constant and the displacement decreases every trial, multiplying the force with the decreasing displacement will result in decreasing work done.

 

 

2.    In part 1, should the power expended be increasing every trial? Explain.

 

The power expended must be constant in every trial since the battery supplies the energy to fan cart. Knowing this, we (our group) tried our best to conserve its energy by not playing with it. Since the results we calculated have a very small difference in each trial, it is safe to say that the power expended is constant.

 

 

 

    Part 2.

 

1.     In Figure 6, why is it incorrect to calculate the work done by multiplying the spring balance reading F and

    the horizontal displacement x?

           

It is incorrect to calculate the work done by multiplying the spring balance reading F and the horizontal displacement because the height of the string is increasing. Multiplying F with the horizontal displacement can give erroneous results because W=Fx is only used when the force and displacement have the same direction. Instead we use the formula W = wL (1-cos Θ) which is equal to the gravitational potential energy (GPE) of the body because the mass is slowly lifted to an increased height (h_f). 

 

 

 

 

 

 

 

 

 

ANALYSIS

 1.  In Table 1, is the work done by the fan cart constant? Why or why not?

              

      No, the work done by the fan cart is not constant because the force of

      the fan cart is constant. When the force of the cart is constant and the

      displacement increases, the work done also increases.        

 

 

 2.  In Table 1, is the power expended by the fan cart constant? Why or why not?

 

     Yes the power expended by the fan cart is constant because the battery is

      the one supplying the energy to the fan cart. Since the battery is the one   

      supplying the energy, the power supplied is constant.

 

 

 3.  In Table 2, how does the work done compare with the increase in gravitational potential energy? Does        

     your result agree with the theory? Why or why not?

 

     When an object is lifted up, its gravitational potential energy is   

     increased. The mass is slowly pulled so that the kinetic energy can be taken as   

     constant, which we did exactly during the experiment.  Thus, from this theory,

     the work done on the curved path is equal to the change in the gravitational

     potential energy. Our results agree with the theory because the calculated

     results of the experimental work done and gravitational potential energy   

     are almost the same.

 

 

CONCLUSION

     1.  What is the correct relationship between the applied force and the work done?

          

          Work done is directly proportional to the applied force in the direction of    

          the displacement. Since work done is equal to the force multiplied by time,

          we can conclude that as the work done is increasing given the displacement

          constant, the force is also increasing.

 

     2.  What is the correct relationship between the displacement and the work done?

 

         Work done is directly proportional to the direction of displacement. Since

           work is equal to force multiplied by time, we can conclude that as the

           displacement is increased given the force constant, the work done is also

           increasing.                                                                                                        

 

 

     3.  What is the correct relationship between the work done and the power expended?

 

         Since power is equal to work over time, given constant time, we can conclude

           that the power expended is directly proportional to the work done.

         

 

 

 

 

 

RESEARCH RELATED TO EXPERIMENT/APPLICATIONS

 

The word work has a variety of meanings. It means different things to different people. To some, mere application of force is already considered as work. But in physics, work is done only when a force is applied to a body and moves it.

 

There are several good examples of work which can be observed in everyday life - a horse pulling a plow through the field, a father pushing a grocery cart down the aisle of a grocery store, a freshman lifting a backpack full of books upon her shoulder, a weightlifter lifting a barbell above his head, an Olympian launching the shot-put, etc. In each case described here there is a force exerted upon an object to cause that object to be displaced.

 

The quantity which has to do with the rate at which a certain amount of work is done is known as the power. Power is the rate at which work is done. It is the work/time ratio. The standard metric unit of power is the Watt (J/s). Another unit of work usually associated with power rating of machines is the Horsepower.

 

Most machines are designed and built to do work on objects. All machines are typically described by a power rating. The power rating indicates the rate at which that machine can do work upon other objects. Thus, the power of a machine is the work/time ratio for that particular machine. A car engine is an example of a machine which is given a power rating. The power rating relates to how rapidly the car can accelerate the car. Suppose that a 40-horsepower engine could accelerate the car from 0 mi/hr to 60 mi/hr in 16 seconds. If this were the case, then a car with four times the horsepower could do the same amount of work in one-fourth the time. That is, a 160-horsepower engine could accelerate the same car from 0 mi/hr to 60 mi/hr in 4 seconds. The point is that for the same amount of work, power and time are inversely proportional. The power equation suggests that a more powerful engine can do the same amount of work in less time.

A person is also a machine which has a power rating. Some people are more power-full than others. That is, some people are capable of doing the same amount of work in less time or more work in the same amount of time.