Block on inclined plane problem

AI Thread Summary
A block is projected up a 22-degree inclined plane with an initial speed of 3.0 m/s, and the acceleration due to gravity is calculated as -3.67 m/s². The distance the block travels up the plane is determined to be 1.23 meters, taking approximately 0.41 seconds to reach its peak. The time to return to the starting point is calculated using kinematic equations, resulting in a total time of 1.64 seconds for the entire trip. The discussion emphasizes the importance of using appropriate kinematic formulas for motion with acceleration, rather than assuming constant speed. The problem showcases the symmetry in motion under constant acceleration, allowing for straightforward time calculations for both ascent and descent.
Cyoder
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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, v^2 = v_0^2 + 2 a \Delta x and a = gsin\theta 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!
 
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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?
 
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