Determine the acceleration as it slides down the plane

[mortho]

[SOLVED] Forces problem

Homework Statement

A block lies on a smooth plane tilted at an angle of 23 degrees to the horizontal. Ignore Friction
1)Determine the acceleration as it slides down the plane.
2) If the block starts from rest 8.90 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline?

Homework Equations

F=ma

The Attempt at a Solution

I couldn't understand the problem at all! how can i find acceleration without any given mass ? PLEASEEE HELP

 

[hage567]

You don't need to know the mass. What is the force causing the acceleration? (you need to take into account the angle of the incline)

Show what you've tried.

 

[mortho]

well the only way i really know how to figure it out is by finding Fg which is done by mass..then finding Force normal and dividing by mass. Could i somehow use the height 8.90 m?

 

[hage567]

Have you drawn a diagram? Label all of the forces on the block. Do you know how to resolve forces into their components? Only a certain amount of gravity is causing the mass to go down the incline. This is where the angle comes in. Once you figure out that, you can use Newton's second law to find the acceleration.

This site might help: http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html#cl

Could i somehow use the height 8.90 m?

Save that for part (b).

 

[mortho]

okay thanks the site did help although i would've never thought of using the equation they did..so i got the acceleration and now need to find Vf which i used Vf=squareroot of 2ax and used squareroot 2*3.83*3.48 ..for the x i used sin 23 (8.90) I'm not too sure that's right but anyway my answer turned out to be 5.16 ..i don't think that seems right is it?

 

[hage567]

okay thanks the site did help although i would've never thought of using the equation they did..so i got the acceleration and now need to find Vf which i used Vf=squareroot of 2ax and used squareroot 2*3.83*3.48 ..for the x i used sin 23 (8.90) I'm not too sure that's right but anyway my answer turned out to be 5.16 ..i don't think that seems right is it?

I'm confused on what you did with x. Are you thinking the block starts at a height of 8.90 m? I don't think that's the what the question is asking.
I take it as the block is sliding 8.90 m along the surface of the incline. So since you know the acceleration of the block in that direction, you would take x = 8.90 m.

 

[mortho]

Ohhh ok..i thought 8.90 was the height so i tried to find the sliding length or hypeteneuse. But i used 8.90 and my answer turned out to be 8.26 ..is that right?

 

[hage567]

Ohhh ok..i thought 8.90 was the height so i tried to find the sliding length or hypeteneuse. But i used 8.90 and my answer turned out to be 8.26 ..is that right?

Well, I guess it could mean the height, sometimes questions are vague on explaining that. But I think it means along the incline. So if that's the way you take it, your answer looks reasonable to me.

If you took it as the height, I think your answer (the 5.16 m/s) is reasonable for that situation.

So either way, you've got it.

 

[mortho]

Thank You Soooooooooooo Much! You Pretty Much Saved My Life!

 

[hage567]

Thank You Soooooooooooo Much! You Pretty Much Saved My Life!

lol you're welcome.

 

[mortho]

Oh wait! if I'm not bothering you enough..could you help me with one more problem..the things' called rope forces...the person who's helping is busy it seems so i don't get much of a response and I'm lost on that.i think that one's supposed to be an easy type problem though.