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

In summary, the problem involves a block being pushed up a frictionless ramp at a constant speed and then being released to slide up the ramp until it stops. The question is asking for the magnitude of the acceleration of the block after it is released. To solve this, one can use the conservation of energy principle and calculate the height the block reaches by converting its kinetic energy to potential energy. Then, using the distance traveled and initial velocity, the acceleration can be calculated using a kinematic equation. Additionally, it is important to resolve the acceleration due to gravity into components parallel to the slope.
  • #1
Eunes
7
0

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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  • #2
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.
 
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  • #3
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...
 
  • #4
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?
 
  • #5
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.
 
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  • #6
You need to resolve the acceleration due to gravity into components, you seek its component parallel to the slope.
 
  • #7
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
 
  • #8
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
 
  • #9
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?
 

1. What is the formula for calculating the magnitude of acceleration on an inclined plane?

The formula for calculating the magnitude of acceleration on an inclined plane is a = g sinθ, where a is the acceleration, g is the acceleration due to gravity (9.8 m/s²), and θ is the angle of the inclined plane.

2. How does the angle of the inclined plane affect the magnitude of acceleration?

The angle of the inclined plane affects the magnitude of acceleration by changing the component of the force of gravity acting on the block along the direction of motion. As the angle increases, the component of the force of gravity acting on the block along the direction of motion decreases, resulting in a smaller magnitude of acceleration.

3. Does the mass of the block have an impact on the magnitude of acceleration?

Yes, the mass of the block does have an impact on the magnitude of acceleration. According to Newton's Second Law, acceleration is directly proportional to the net force acting on an object and inversely proportional to its mass. Therefore, a larger mass will result in a smaller magnitude of acceleration.

4. Can the magnitude of acceleration ever be greater than the acceleration due to gravity?

No, the magnitude of acceleration can never be greater than the acceleration due to gravity. This is because the acceleration due to gravity is a constant value (9.8 m/s²), and the magnitude of acceleration on an inclined plane is always equal to or less than this value.

5. Is the magnitude of acceleration the same throughout the motion of the block on an inclined plane?

No, the magnitude of acceleration is not the same throughout the motion of the block on an inclined plane. At the beginning of the motion, the magnitude of acceleration is higher as the block overcomes the force of static friction. As the block moves up the inclined plane, the magnitude of acceleration decreases due to the decrease in the component of the force of gravity along the direction of motion.

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