Physics Questions: Frictionless Incline + Pendulum Speed

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The discussion revolves around two physics questions related to conservation of energy. The first question involves a frictionless incline where a box slides from a height of 4.0m, with calculations leading to a final speed of 13.3m/s at the bottom. The second question pertains to a pendulum raised to 0.25m, resulting in a speed of 2.2m/s at the equilibrium position. Participants emphasize that the slope does not affect the energy conservation principle since potential energy (PE) converts to kinetic energy (KE) without losses. The concept of a "conservative field" is also mentioned for further understanding of energy conservation.
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Hello all, just started taking physics 30, or pyhsics 12 and need some help with two questions. Any help is appreciated.

1. A frictionless incline with a vertical rise of 4.0m and a slope of 9.0m is used to slide a box from the 4.0m height. If the box starts from rest at the top of the incline what is its speed at the bottom.

2. A 1.0m pendulum is raised to a 0.25m height above the equilibrium position. What is the speed of the pendulum bob as it passes through the equilibrium position.

Yea probably easy questions and supposed to be review but having a hard time with them, lol. Thanks for any help.
 
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Both questions are to do with conservation of energy (surprise!).

Try not to get caught up in thinking it's more complex than this - Ask yourself what kind of energy they have at the start of the experiment and what they have at the point of measurement (no friction so no losses to worry about).

Formulate it in a mathematical expression and the answer you require will drop out.
 
Last edited:
hrm

Well I come up with the equation v2 = 2gh , but since there is a slope of 9m in the 4 meter plane dosent that change anything.
 
The slope won't change anything because, in the absence of friction, PE + KE = constant (as you have rightly used in your calculation). So it's the magitude of the change in PE that affects your KE at the end, not how you get there.

It's a property a "conservative field" which you may want to look at if you need a deeper understanding.
 
Thanks =)

Thanks.
v2 = 2gh
v2 = 2(9.81)(9.0m)
v2= 176.58m/s
v = 13.3m/s

v2 = 2gh
v2 = 2(9.81)(0.25m)
v2 = 4.905
v = 2.2m/s

Is what I ended up with. Least I am on the right track I hope. =)

Edit: Also going to look up what you suggested "conservative field"
 
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