Inclined Plane Problem Two Parts

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Homework Help Overview

The discussion revolves around a physics problem involving a block sliding down an inclined plane at an angle of 24.5° to the horizontal, with two parts: determining the acceleration of the block and calculating its speed when it reaches the bottom after starting from rest at a height of 9.50 m. The problem is set within the context of kinematics and dynamics, specifically addressing forces on an inclined plane.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of acceleration using trigonometric functions and Newton's second law. Questions arise regarding the correctness of the initial calculations and the interpretation of forces acting on the block. There is an exploration of the relationship between gravitational force components and the incline.

Discussion Status

The discussion is active, with participants questioning the initial approach to calculating acceleration and suggesting a need for clarification on force components. Some guidance has been offered regarding the application of Newton's laws and the correct interpretation of forces in the context of the incline.

Contextual Notes

Participants express confusion about the application of trigonometric functions in their calculations and the interpretation of the forces involved. There is a mention of a resource for further understanding of vector components, indicating a need for deeper exploration of the concepts at play.

sugarntwiligh
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Homework Statement


The block shown in Fig. 4-48 lies on a smooth plane tilted at an angle θ = 24.5° to the horizontal. Ignore friction.

http://www.webassign.net/giancoli5/4_48.gif (visit this site for picture)

(a) Determine the acceleration of the block as it slides down the plane.

(b) If the block starts from rest 9.50 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline?


The Attempt at a Solution



a.)
9.8/sin(24.5)=23.6 m/s^2 which is the acceleration down the slope where the x=ma and the y=mg. Is this correct? It feels like I am forgetting something...
b.)
From the equation v^2=vi^2+2ad,
v^2=0+2*23.6*9.5
v=21.2m/s at the bottom of the incline.
 
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sugarntwiligh said:
a.)
9.8/sin(24.5)=23.6 m/s^2 which is the acceleration down the slope where the x=ma and the y=mg. Is this correct?
No, it's not. What forces act in the x direction? Apply Newton's 2nd law. (Sanity check: Compare your calculated acceleration with g. Does your answer make sense?)
b.)
From the equation v^2=vi^2+2ad,
v^2=0+2*23.6*9.5
v=21.2m/s at the bottom of the incline.
If you had the correct acceleration, that approach would work.
 
Doc Al said:
No, it's not. What forces act in the x direction? Apply Newton's 2nd law. (Sanity check: Compare your calculated acceleration with g. Does your answer make sense?)

If you had the correct acceleration, that approach would work.

I didn't use any value of x in my equation. But I am still confused about why its wrong, because sin(24.5)=(9.8m)/(ma). Where mg is the opposite and ma is the hypotenuse. And then solving for a gives me the same answer.

My answer would make more sense if I did 9.8sin(24.5)=4.06, and I think this is the right way to go, but why?
 
sugarntwiligh said:
I didn't use any value of x in my equation. But I am still confused about why its wrong, because sin(24.5)=(9.8m)/(ma). Where mg is the opposite and ma is the hypotenuse. And then solving for a gives me the same answer.
To use a triangle to find force components, you must realize that the total force is "mg", so mg must be the hypotenuse of your right triangle. (That right triangle is not simply the incline.)

My answer would make more sense if I did 9.8sin(24.5)=4.06, and I think this is the right way to go, but why?
You need to learn how to find the components of the weight parallel and perpendicular to the incline surface. Read this: http://www.glenbrook.k12.il.us/gbssci/Phys/Class/vectors/u3l3e.html"
 
Last edited by a moderator:
Ohhhhhhhhh I get it! Thanks!
 

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