What is the Velocity and Distance of a Block on an Incline?

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A block of mass 2.00 kg is released from a height of 0.300 m on a 50.0° incline, and the goal is to determine its velocity upon leaving the incline, the distance it travels before hitting the floor, and the time taken to reach the ground. The initial attempt calculated an acceleration of 7.51 m/s² and a time of 0.308 seconds, but the resulting velocity was incorrect. The discussion suggests using a single constant acceleration equation or conservation of energy for a more accurate solution. The hypotenuse of the incline is noted to be 0.357 m. The conversation emphasizes the importance of correct equation usage in solving the physics problem.
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Homework Statement


A block of mass m = 2.00 kg is released from rest at h = 0.300 m from the surface of a table, at the top of a θ = 50.0° incline as shown below. The frictionless incline is fixed on a table of height H = 2.00 m.

http://www.webassign.net/pse/p5-62.gif"

(b) What is the velocity of the block as it leaves the incline?


(c) How far from the table will the block hit the floor?


(d) How much time has elapsed between when the block is released and when it hits the floor?


The Attempt at a Solution



b. ax= 7.51m/s2
Hypothenuse: 0.357m
time to travel 0.357m:
d= Vot +1/2at2
t=\sqrt{2da}
t= 0.308s

Vx=Vxi + at
Vx= (7.51)(0.308)= 2.31 m/s
This was the wrong answer. I thought that finding the velocity at the time at which the block travels the hypothenus would give the correct velocity.

c. Since my velocity is wrong I can't solve for c or d.

c.
 
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Welcome to PF!

Hi xjasonx1! Welcome to PF! :wink:
xjasonx1 said:
Hypothenuse: 0.357m

Nooo … I think you've used tan instead of … ? :smile:

(and why have you used two equations? … there's a single constant acceleration equation that will do the trick … or you could even use conservation of energy!)
 
thanks tim!
 
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