Homework Help: PHYSICS

Posted by Sarah on Saturday, January 30, 2010 at 4:00pm.

A pendulum was set up and measurements were made to enable the mechanical energy to be calculated at the start position S and the lowest point of the pendulums swing L.

The mass of the pendulum bob was determined on an electronic scale and its diamter was measured using calipers. The initial height was measured with a meter stick. At the lowest point of its swing, the pendulum bob broke a photogate light beam. The time interval that the light was interrupted was recorded on an electronic timer attached to the photogate.
Use the following data to complete a report.
MAss of pendulum bob = 240.3 g
Diameter of pendulum bob = 3.50 cm
Initial height of pendulum bob = 48.0 cm
Length of pendulum string = 2.14 m
Time interval of photogate light interruption = 11.8 ms

Your report should include the following:
(a) conclusion as to whether or not the pendulum demonstrated the law of conservation of energy
(b) calculations of the efficiency of the pendulum as a mechanical machine

I am really confused how to even start this question. I think that I should figure out the Emechanical = Ek + eg

=1/2(240.3)(0) + 240.3(9.8)(.48)
=1130.37J

but have no idea how to calculate the speed of when it hit. I don't know where to go from here at all, please help!

PHYSICS - Damon, Saturday, January 30, 2010 at 4:10pm
The clue is the diameter of the bob
it broke the beam for 11.8 *10^-3 seconds
it is .035 meters in diameter
so it went
.035 meters in 11.8^10^-3 seconds
which is about 2.97 meters/s
PHYSICS - Damon, Saturday, January 30, 2010 at 4:14pm
(1/2) 240.3 v^2 = gh = 1130.37
v^2 = 9.407
v = 3.067 ideally but we only measured 2.97 m/s so some energy got lost along the way like air friction and stuff.
typo, left m out - Damon, Saturday, January 30, 2010 at 4:16pm
(1/2) 240.3 v^2 = mgh = 1130.37
v^2 = 9.407
v = 3.067 ideally but we only measured 2.97 m/s so some energy got lost along the way like air friction and stuff.

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