Inclined Plane Problem Two Parts

AI Thread Summary
The discussion revolves around solving an inclined plane physics problem involving a block sliding down a frictionless surface at an angle of 24.5°. The initial calculation for acceleration was incorrectly derived, leading to confusion about the application of Newton's second law. The correct approach involves resolving the gravitational force into components parallel and perpendicular to the incline. The final speed of the block at the bottom of the incline was calculated using the incorrect acceleration, prompting a need to reassess the force components. Understanding how to properly calculate these components is crucial for solving the problem accurately.
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"
 
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Ohhhhhhhhh I get it! Thanks!
 
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