# Magnitude of the acceleration of the block after it is released up an inclined plane

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1. Oct 22, 2014

### Eunes

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

2. Relevant equations
F=ma?
F(cos(theta))?

3. The attempt at a solution
The answer is 4.9 m/s^2 for example 1,

2. Oct 22, 2014

### Staff: Mentor

Hi Eunes. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

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: May 7, 2017
3. Oct 22, 2014

### NTW

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. Oct 22, 2014

### Eunes

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. Oct 22, 2014

### Eunes

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: May 7, 2017
6. Oct 22, 2014

### Staff: Mentor

You need to resolve the acceleration due to gravity into components, you seek its component parallel to the slope.

7. Oct 22, 2014

### NTW

'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. Oct 22, 2014

### Eunes

So what exactly is the formula to calculate it please?

9. Oct 22, 2014

### Eunes

Right, so m*g*sin(theta) is the parallel component. What would you plug in, and what did you get for the answer?

10. Oct 22, 2014

### NTW

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. Oct 22, 2014

### Staff: Mentor

"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?