What is the magnitude of acceleration for a block released up an inclined plane?

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Homework Help Overview

The discussion revolves around the acceleration of a block released on a frictionless inclined plane after being pushed up at a constant speed. Two examples are presented, each involving a block of different masses and angles of inclination. Participants are exploring the relationship between kinetic energy, potential energy, and the forces acting on the block as it moves along the ramp.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of energy conservation principles, considering kinetic and potential energy. There are inquiries about the distinction between speed and velocity, and how to apply trigonometric functions to resolve forces. Some participants express uncertainty about the initial conditions and seek clarification on the relevant formulas.

Discussion Status

The discussion is active, with participants providing hints and guidance on how to approach the problem. There is a focus on understanding the components of forces acting on the block and the implications of energy conservation. Multiple interpretations of the problem are being explored, particularly regarding the application of trigonometric functions and the nature of acceleration.

Contextual Notes

Participants are navigating through the complexities of the problem, including the need to clarify definitions and assumptions about energy and motion. There is an acknowledgment of the distinction between speed and velocity, which may impact the understanding of the problem.

Eunes
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Homework Statement


1st example:

A 10 kg block is pushed up a ramp at a constant speed of 5.7 m/s. The ramp makes an angle of 30 degrees to the horizontal and the ramp is frictionless. The block is released and allowed to slide up the ramp until it stops. What is the magnitude of the acceleration of the block after it is released? /////

2nd example:

A 14 kg block is pushed up a ramp at a constant speed of 6.5 m/s. The ramp makes an angle of 25 degrees to the horizontal and the ramp is frictionless. The block is released and allowed to slide up the ramp until it stops. What is the magnitude of the acceleration of the block after it is released?

Homework Equations


F=ma?
F(cos(theta))?

The Attempt at a Solution


The answer is 4.9 m/s^2 for example 1,[/B] but how do you get that answer? Plus, how to solve example 2 and what is the answer? (Please help, step by step for dummies)
 
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Hi Eunes. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Draw a diagram of what you are dealing with.

You know about sine, cosine, tangent, etc.?

Hint: acceleration is a vector.
 
Last edited by a moderator:
A nice problem to be solved 'by energies':

The KE of the block is, in the instant that you release it is easy to calculate, as you know its mass and velocity. Once released, the block decelerates while climbing some distance up the ramp, till all that its KE is converted to PE with respect to the height of the point where you released it.

What's that height h? Well you know that energy is conserved In this case, PE = KE => mgh = KE. It's easy to solve for h, and once you have got h, as you know the angle of the ramp, it's easy to calculate the distance that the block moved up that ramp...

That distance found, remember that the block decelerates uniformly from the moment it is released till the moment it stops. It's a case of negative acceleration. You have the initial velocity and the distance that the block has moved. You must know an equation that, plugging in it acceleration and distance, gives you the velocity. Use it, and solve for acceleration, plugging distance and velocity...
 
NTW said:
A nice problem to be solved 'by energies':

The KE of the block is, in the instant that you release it is easy to calculate, as you know its mass and velocity. Once released, the block decelerates while climbing some distance up the ramp, till all that its KE is converted to PE with respect to the height of the point where you released it.

What's that height h? Well you know that energy is conserved In this case, PE = KE => mgh = KE. It's easy to solve for h, and once you have got h, as you know the angle of the ramp, it's easy to calculate the distance that the block moved up that ramp...

That distance found, remember that the block decelerates uniformly from the moment it is released till the moment it stops. It's a case of negative acceleration. You have the initial velocity and the distance that the block has moved. You must know an equation that, plugging in it acceleration and distance, gives you the velocity. Use it, and solve for acceleration, plugging distance and velocity...

But I don't know the velocity -- I know the speed.
I am pretty sure I have to find the acceleration as soon as it is let go, so we don't have to go too in depth.

I'm sorry for my lack of knowledge, but what was KE/PE stand for?
 
NascentOxygen said:
Hi Eunes. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Draw a diagram of what you are dealing with.

You know about sine, cosine, tangent, etc.?

Hint: acceleration is a vector.

Thank you for the rapid reply -- yes, I do know the trigonometric functions and I do know that acceleration is a vector, but how do I implement that knowledge into this problem? Alternatively, what are the first few steps/an equation that I can use to find this answer?

Thanks again.
 
Last edited by a moderator:
You need to resolve the acceleration due to gravity into components, you seek its component parallel to the slope.
 
Eunes said:
But I don't know the velocity -- I know the speed.
I am pretty sure I have to find the acceleration as soon as it is let go, so we don't have to go too in depth.

I'm sorry for my lack of knowledge, but what was KE/PE stand for?

'Velocity' and 'speed' are distinctions that exist -so far as I know- only in English. It's the same, but seen as a vector or as a scalar.

KE = kinetic energy

PE = potential energy
 
So what exactly is the formula to calculate it please?
NTW said:
'Velocity' and 'speed' are distinctions that exist -so far as I know- only in English. It's the same, but seen as a vector or as a scalar.

KE = kinetic energy

PE = potential energy
 
NascentOxygen said:
You need to resolve the acceleration due to gravity into components, you seek its component parallel to the slope.
Right, so m*g*sin(theta) is the parallel component. What would you plug in, and what did you get for the answer?
 
  • #10
Eunes said:
So what exactly is the formula to calculate it please?

You shouldn't ask for 'the formula', as if it were a magic solution: getting the right formula, plugging in it the values, and solving the problem...

Formulas are useful, but -first of all- you have to understand the problem, and then devise a strategy to solve it.
 
  • #11
Eunes said:
Right, so m*g*sin(theta) is the parallel component. What would you plug in, and what did you get for the answer?
"m" does not belong in the expression for acceleration. We are simply looking for that component of gravity, g acting along the slope.

What do you get for the answer?
 

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