Calculate the mechanical energy

[8uhohs]

Thanks in advance for the help! I really appreciate it :D

Homework Statement

A pendulum was swung once and when the pendulum bob was at the lowest point of its swing, it broke a photogate light beam. The following information was collected:

• Diameter of pendulum bob = 3.50cm = 0.035m
• mass of pendulum bob= 240.3g = 0.2403kg
• initial height of pendulum bob = 48.0cm = 0.48m
• length of pendulum string = 2.14m
• time interval of photogate light interruption = 11.8ms = 0.0118s

Calculate the mechanical energy at the start position,S, and the lowest point of the pendulum's swing,L.

Homework Equations

v_L=\frac{d}{t} , where d is the diameter and t is the time
E_mechanical=\frac{1}{2}mv²+mg\Deltah

The Attempt at a Solution

yea I'm kind a positive this is not right but here's the only thing that i can come up with...

v_L=\frac{d}{t}
v_L=\frac{0.035}{0.0118}
v_L=2.97m/s

at L,
E_mechanical=\frac{1}{2}(0.2403kg)(2.97m/s)²+(0.2403kg)(9.8m)(0)
I'm guessing the height of the pendulum bob at it's lowest is 0m
E_mechanical= 1.06J

and for S, I'm guessing it has something to do with the law of conservation of energy...

yea i think I'm wrong since i didn't use some of the information given and later on it tells me to make a conclusion as to whether or not the pendulum demonstrated the law of conservation of energy ><

 

[kuruman]

The lowest point of the motion is the equilibrium position when the pendulum just hangs at rest. Is the photobeam blocked or unblocked at that point?

 

[8uhohs]

The lowest point of the motion is the equilibrium position when the pendulum just hangs at rest. Is the photobeam blocked or unblocked at that point?

the photobeam should be blocked at that point and the question also give a diagram and the bob does continue on after reaching the lowest point, so it doesn't just hang at rest

 

[kuruman]

I know that. So what is Δh from the point when the photogate is blocked for the first time until the pendulum reaches the lowest point?