Procedure M=270 g (hanger +220g); Tape the 220 g to the hanger that is 50g m=250g (hanger +200g);...

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Physics

Procedure

M=270 g (hanger +220g); Tape the 220 g to the hanger that is50g

m=250g (hanger +200g); Tape the 200 g to the hanger that is50g

Set the apparatus so that when m is on the floor, M is at heighth=1.50m; Make sure you measure and record the actual value of hthat you use.

Release the system from rest and measure the time taken for m1to fall on the floor.

SAFETY – When M hits the floor it becomes a free projectile;have a group member catch it!!!!

You will perform 5 trials and record the data in thefollowing

                

Table1:                                                                                                                                                         

Trial #

Time, second

1

5.55

2

5.43

3

5.71

4

5.38

5

5.66

Data and Calculation

Determine the average time tav from your data in Table 1.

In order to determine the final velocity of mass m1 just beforeit hits the floor you can use:

Method 1:

It can be shown that the final velocity of the mass m1is givenby eq. (1):

h=12(vi+vf)t eq. (1)

Method 2:

Since mass M started from rest we could also use

vf=vi+at eq. (2)

The following equation for the acceleration of the Atwoodmachine can be derived using FBD:

a=M-mM+mg eq. (3)

Calculating Energy

Calculate the final kinetic energy KEf , and the final potentialenergy PEf using the following equations:

KEf=12(M+m)vf2 eq. (4)

PE=mgh                   eq. (5)

The final total energy is given by:

Ef=Kf+PEf            eq. (6)

(Note: mass needs to be in kilograms if you are reporting theenergy in units of Joules).

Record the final velocity, obtained using Method 1 and Method 2in Table 2 below.

Table 2:

Method #

vf, m/s

1

2

Using the equations above, compute the initial and final totalmechanical energy of the system for each of the final velocities(from Method 1 and Method 2) and record your results in Table 3below.

Table 3:

Method #

Ei=PEi+Ki

(J)

Ef=PEf+Kf

(J)

Percentage difference

1

2

If total energy is conserved, then Ei=Ef. Find the percentagedifference between the two measured values of Ei and Ef, using thefollowing formula:

%difference = Ei-EfEi+Ef2×100

Analysis:

Answer the following questions:

Based on your experiment would you conclude that the mechanicalenergy was conserved?

How much (if any) energy was lost (or gained)?

Where did the energy go or come from?

What are the sources of error?

Which of the two methods used to compute the final velocity isbetter suited for this experiment and why?

Based on your knowledge of principle of conservation ofmechanical energy when forces are conservative, answer thefollowing conceptual questions using the words - increase,decrease, or        remains the same.Refer to figure 1. The gravitational potential energy referencelevel will be the ground as indicated by y=0.

As m rises, what happens to its gravitational potentialenergy?__________

As M falls, what happens to its gravitational potential energy?__________

As the masses move what happens to the total gravitationalpotential energy of the system? __________

As M falls, what happens to its kinetic energy? __________

As m rises, what happens to its kinetic energy? __________

As the masses move what happens to the total kinetic energy ofthe system? __________

Assuming no friction, as m rises and M falls, what happens tothe total mechanical energy of the system? __________

Determine your own power rating by measuring the time it takesyou to climb a flight of stairs. Ignore the gain in kineticenergy.   Do 5 trials and submit the table below.   Outline the strategy for obtaining your data andprovide the calculations you used to determine you power (hint:refer to example 7.11 in the text).

Trial #

Time, second

1

2

3

4

5

Answer & Explanation Solved by verified expert
4.2 Ratings (693 Votes)
AnalysisAnswer the following questionsBased on your experiment would you conclude that the mechanicalenergy was conservedyes we can say that the mechanical energy is conservedbecause in method1 it decreased and in method2 it increased whichis due to the errorHow much if any energy was lost or gainedmethod1 022 joules was    See Answer
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