# Block on inclined plane problem

1. Sep 8, 2010

### Cyoder

1. The problem statement, all variables and given/known data
A block is given an initial speed of 3.0m/s up a 22 degree plane.

a.) How far up the plane will it go?

b.) How much time elapses before it returns to the starting point?

no friction btw

2. Relevant equations
Assuming that https://www.physicsforums.com/showthread.php?t=427167&highlight=cart+pushed+up+plane" is correct, $$v^2 = v_0^2 + 2 a \Delta x$$ and a = gsin$$\theta$$ should work.

3. The attempt at a solution
a=-9.8sin(22)

a=-3.67 m/s^2

initial velocity = 3.0 m/s of course

so

0 = 9 - 7.34 delta X

a.) delta X = 1.23m

t = 1.23 / 3.0

t = .41 seconds

So the trip up is .41 seconds. How do I get the time it takes to roll back?

I'm bad with inclined plane problems, mostly because they scare me :yuck:. What do I do from here, assuming I haven't made a major error up to now?

Also, every time I try to preview my message, it adds a new set of instructions. Very irritating!

Last edited by a moderator: Apr 25, 2017
2. Sep 8, 2010

### cepheid

Staff Emeritus
Welcome to PF

The step in red is definitely wrong. You can't just say time = distance/speed, because this equation is only correct for motion at a constant speed. The block is not moving at a constant speed -- it is accelerating. You will have to use some of the other kinematics formulas you have learned for motion with constant acceleration -- ones that include time as a variable.

As for your second question: there is a nice symmetry in the problem. The time it takes to slow to a stop over a given distance should be the same as the time it takes to speed back up over that distance, since the acceleration is the same.

3. Sep 8, 2010

### Cyoder

Ok thank you cepheid!

i have selected V = u + at

0 = 3 - 3.67t

t = 0.82 sec

From your last statement, would it be correct to say the the total time is 0.82 x 2?

so Total time = 1.64 sec?