What is the speed of the rock physics

In summary, the problem involves a small rock of 0.240 kg being released from rest at point A, on the top edge of a large, hemispherical bowl with a radius of 0.500 m. The rock is treated as a particle and slides instead of rolling. The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl is -0.160 J. The formula used to solve the problem is mgh=mv^2/r, and the final answer is a speed of 2.94 m/s. Energy conservation is also used in the solution, with the equation being Kinetic energy(final)+potential energy(final)+Work=potential energy inital+
  • #1
melodrameric
7
0
Again another work/kinetic energy problem I need help with.
A small rock with a mass of 0.240 kg is released from rest at point A, which is at the top edge of a large, hemispherical bowl with a radius R = 0.500 m. Assume that the size of the rock is small compared to the radius of the bowl, so that the rock can be treated as a particle, and assume that the rock slides rather than rolling. The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl is - 0.160 J. What is the speed of the rock when it reaches point B?
I really don't know where to start.
picture is attached.
 

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  • #2
Use energy methods.

Point A:

Whats the associated energy at that point a? Gravitational? Kinetic? Electric?

Whats the associated energy at point b?

Show some work and well go from there.

Thats a very nice picture by the way, how did you get it?
 
Last edited:
  • #3
hey i am not sure abot the answer. is this correct

mgh=mv^2/r
(.24*10*.5)-.16=(.24*v^2)/2
v=2.94m/s
 
  • #4
It is very difficult to help you when you won't at least indicate what you DO know! Why do you think that formula will help you? Are you using kinetic energy and potential energy? But what about "The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl is - 0.160 J"?
 
  • #5
i hav reduced it from the potential energy. so it equalls to the kinetic energy
 
  • #6
mgh=mv^2/r...

So PE = Fc?

Fc being the centripetal force. I don't think that's right. PE is in joules and Fc is in Newtons.

I think it might actually be:

PE=KE

mgh = [(1/2) mv^2] + Wf,

where height (h) is your radius and W is the work done by friction (-.160J). When you use -.160J make it positive in the equation since the KE plus the work done by friction should be equal to the PE.

Solve for v.
 
  • #7
thanx for pointing out my mistakes.
 
  • #8
I would approach the problem using energy as well. Since there is frictional force then energy conservation must involve work. so Kinetic energy(final)+potential energy(final)+Work=potential energy inital+kinetic energy initial. Where potential initial=mgR and kinetic initial=potential final=0 and kinetic final=.5mv^2. I am not entirely sure someone correct my mistakes ^_^
 

1. What is the definition of rock physics?

Rock physics is the study of how rocks and other geological materials respond to various physical forces, such as pressure, temperature, and stress.

2. How is the speed of rock physics measured?

The speed of rock physics is typically measured using seismic wave analysis, which involves sending sound waves through rocks and measuring their travel time and amplitude.

3. What factors affect the speed of rock physics?

The speed of rock physics is affected by several factors, including the type of rock, its composition and structure, temperature, pressure, and the presence of fluids or cracks within the rock.

4. Why is the speed of rock physics important in geology?

The speed of rock physics is important in geology because it can provide valuable information about the properties and behavior of rocks, which is crucial for understanding geological processes and predicting potential hazards such as earthquakes or landslides.

5. How does the speed of rock physics change with depth?

The speed of rock physics generally increases with depth, as the weight of the overlying rocks and the pressure from the Earth's crust compresses the rocks and makes them more compact, resulting in faster seismic velocities.

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