Block on inclined plane problem

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SUMMARY

The inclined plane problem involves a block with an initial speed of 3.0 m/s moving up a 22-degree incline without friction. The acceleration is calculated as -3.67 m/s², leading to a maximum distance of 1.23 m up the plane. The time taken to reach this height is 0.41 seconds. The total time for the block to return to the starting point is determined to be 1.64 seconds, based on the symmetry of the motion and the use of kinematic equations.

PREREQUISITES
  • Understanding of kinematic equations, specifically for motion with constant acceleration.
  • Knowledge of trigonometric functions, particularly sine for angle calculations.
  • Familiarity with the concept of acceleration due to gravity (g = 9.8 m/s²).
  • Ability to solve basic physics problems involving inclined planes.
NEXT STEPS
  • Study the derivation and application of kinematic equations for uniformly accelerated motion.
  • Learn about the effects of friction on inclined plane problems.
  • Explore the concept of energy conservation in inclined plane scenarios.
  • Practice additional problems involving different angles and initial velocities on inclined planes.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and inclined plane problems, as well as educators seeking to enhance their teaching methods in this area.

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, [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!
 
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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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