# Calculate the mechanical energy

1. Aug 20, 2010

### 8uhohs

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

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

vL=$$\frac{d}{t}$$ , where d is the diameter and t is the time
Emechanical=$$\frac{1}{2}$$mv2+mg$$\Delta$$h

3. 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...

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

at L,
Emechanical=$$\frac{1}{2}$$(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
Emechanical= 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 ><

2. Aug 21, 2010

### 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?

3. Aug 22, 2010

### 8uhohs

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

4. Aug 22, 2010

### 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?