Applying Newton's Laws Projected Up an Incline

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SUMMARY

The discussion focuses on a physics problem involving a block projected up a frictionless incline at an initial speed of 3.43 m/s and an angle of 32.5°. The calculations for distance traveled up the incline yield 2.31 m, while the time taken to reach this height is approximately 0.674 seconds. The final speed of the block upon returning to the bottom is determined using the equations of motion, specifically vf² = vi² + 2ad and vf = vi + at, confirming that the block will return to its initial speed of 3.43 m/s in the opposite direction.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Familiarity with kinematic equations
  • Knowledge of trigonometric functions, specifically sine
  • Concept of frictionless inclined planes
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn about the effects of friction on inclined planes
  • Explore advanced topics in dynamics, such as energy conservation in motion
  • Investigate real-world applications of Newton's Laws in engineering
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Students of physics, educators teaching mechanics, and anyone interested in understanding motion on inclined planes will benefit from this discussion.

djester555
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Projected Up an Incline A block is projected up a frictionless inclined plane with initial speed v1 = 3.43 m/s. The angle of incline is θ = 32.5°.
(a) How far up the plane does it go?

(b) How long does it take to get there?

(c) What is its speed when it gets back to the bottom?



What i got soo far
v * T = d

a) 3.43 * .674 = 2.31 m


b)vi = 3.43
vf = 0
a = - 9.80 sin32.5
t = ?

t = (vf - vi) / a
t = 3.43 / (9.80 sin32.5)
t = 0.674




c) don't even know where to begin on this one
 
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You can solve the problem using the equations of motion

vf2 = vi2 +2ad ---------(1)
vf = vi + at --------(2)
where vf and vi are the final and initial velocity respectively and a is the acceleration, t is the time

Using equation (1), you can solve for d
and using (2) you can solve for t

For the third question, you have to consider the distance the block has traveled
and the block would have this length of 'track' to go back to the bottom.

So, again you can use (1) to solve for vf
(which must equal to 3.43m/s, to the opposite direction)

i don't get what is 'T' used for question a
your working for b is correct but i get 0.651s as my answer, using the same working
 

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