# Inclined Plane Problem Two Parts

1. Oct 12, 2008

### sugarntwiligh

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

3. 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.)
v^2=0+2*23.6*9.5
v=21.2m/s at the bottom of the incline.

2. Oct 12, 2008

### Staff: Mentor

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.

3. Oct 12, 2008

### sugarntwiligh

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?

4. Oct 12, 2008

### Staff: Mentor

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.)

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" [Broken]

Last edited by a moderator: May 3, 2017
5. Oct 12, 2008

### sugarntwiligh

Ohhhhhhhhh I get it! Thanks!