Block on inclined plane problem

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 23K views
Cyoder
Messages
2
Reaction score
0

Homework Statement


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

Homework Equations


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

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 . 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:
Physics news on Phys.org
Welcome to PF

Cyoder said:

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 . 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!

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.
 
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?