What Are the Forces and Energy Involved in a Block Sliding Down a Ramp?

AI Thread Summary
The discussion revolves around a physics problem involving a block sliding down a frictionless ramp. Key points include calculating the normal force at the bottom of the ramp, which is equal to the weight of the block, and at point B, where the normal force accounts for the component of gravity, resulting in a different value. The speed of the block at the end of the ramp is derived from energy conservation principles, yielding a speed of 8.34 m/s. The participant expresses difficulty in determining the launch angle for projectile motion after leaving the ramp and seeks clarification on whether a 45-degree angle can be assumed. Additionally, there is a discussion about the shape of the potential energy graph, which is thought to resemble the ramp's profile.
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1. Homework Statement

A block of mass m=1.62 kg slides down a frictionless surface (see below). The block
is released at a height h=3.91 m from the bottom of the loop. Assume the bottom of the
ramp indicated between the dashed lines in the diagram is a segment of a circle with R = 1.5m. Neglect air resistance.

a) What is the force of the inclined track on the block at the bottom (point A)?
b) What is the force of the track on the block at point B?
c) At what speed does the block leave the track?
d) How far away from point A does the block land on level ground?
e) Sketch the potential energy U(x) of the block. Indicate total energy on the sketch.


2. Homework Equations

Are explained below

3. The Attempt at a Solution

A) I think, it's just the opposite force to the blocks weight at point a, and with no angles, it's simply mg = 15.89 N

B) the normal force at point b now has to take into account the component of gravity in the y direction so would the force the track exert on the block be mgsin45? If so the answer is 10.32 N

C) the speed at the end of the ramp corresponds to the kinetic energy it gained, from the potential energy at the top = mgh. however h in this case = (h - distanceAB). with a little bit of algebra I worked out the distance from the top to point B in the y direction (so new h) is 3.55 m. so PE mgh = 56.36 J = 1/2mv2. rearranging yields v = 8.34 m/s

D) this is where I'm having problems. I know for a normal projectile question, if I'm given a projectile angle, distance from the ground and initial veloctiy I can calculate time in air, and eventually distance travelled. I have 2 of the 3, and cannot find a launch angle. Keep in mind this is not a straight ramp I'm dealing with, so I didn't think trig could help me. Do you think it is safe to just assume launching at a 45 degree angle? Please help if there is a way to do it.

E) I think the potential energy graph looks a lot like the ramp itself, except it starts at the origin and jumps up immediately after. Am I right here?

I only needed help on D) and E), but please even if I'm right and you know I am, confirm my other answers, it will help a lot for the learning process. Thanks in advance
 

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Hi captain Evil ^^

I faced this problem before and i kept the answer in my Computer ^^

Download it from here :

http://www.mediafire.com/download.php?zw11mdmj1kn

See Ya
 
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I don't have word. Can you just describe how you went about part a & b?
 
Wooo Wait friend I faced this problem maybe 2 years ago & actually forgot the whole concept of the problem. So please forgive me, try to read it & understand it by yourself & if U stuck In somewhere tell the Members to clear 4 U that part.

Sorry Again Captain.
 
I have attached the answers for parts A & B .
 

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