Maximum distance traveled?

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  • #1
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A 4.0 kg block has a speed of 9.00 m/s at X. What is the maximum distance, d, traveled by the block. Ignore friction. There is a angle of 35 degrees.

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I know this a very simple problem, and I have solved problems similar to this in my honors physics class; however, I have not solved problems that involve finding the maximum distance traveled. Teacher pulled this from the AP Physics B question bank, but our textbook doesn't address how to solve it. Could someone lend a helping hand? Thanks!
 

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  • #2
LowlyPion
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A 4.0 kg block has a speed of 9.00 m/s at X. What is the maximum distance, d, traveled by the block. Ignore friction. There is a angle of 35 degrees.
___
I know this a very simple problem, and I have solved problems similar to this in my honors physics class; however, I have not solved problems that involve finding the maximum distance traveled. Teacher pulled this from the AP Physics B question bank, but our textbook doesn't address how to solve it. Could someone lend a helping hand? Thanks!
What other information is there? Is there a picture? Is it a block sliding down an incline? Is the block always constrained to have been above the point X
 
  • #3
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There is a picture; apologies for not describing it previously.

There is an incline of 35 degrees with a box of 4.0kg almost to the end of it. The box was previously up the incline quite a bit (distance "d") as indicated by a box with broken lines. Apologies for not providing a picture. Is this description succinct enough to help you out?

Many thanks.
 
  • #4
LowlyPion
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There is a picture; apologies for not describing it previously.

There is an incline of 35 degrees with a box of 4.0kg almost to the end of it. The box was previously up the incline quite a bit (distance "d") as indicated by a box with broken lines. Apologies for not providing a picture. Is this description succinct enough to help you out?

Many thanks.
That makes it solvable then.

Think about how far up the incline you can release it. And what your initial velocity must be to be as far away as possible.
 
  • #5
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I thought about what you wrote and this is what I have figured:

[tex]\Sigma[/tex]F = 9 - 0 = 4(acceleration)
acceleration = 4/9 m/s/s

(final velocity)^2 = (initial velocity)^2 + 2(acceleration)(distance)

(9)^2 = (0)^2 + (2)(4/9)d

d=91.125 m

____

However, I am almost certainly incorrect. I never took into account the incline of the angle (35 degrees), so I think my logic is flawed.
 
  • #6
LowlyPion
Homework Helper
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I thought about what you wrote and this is what I have figured:

[tex]\Sigma[/tex]F = 9 - 0 = 4(acceleration)
acceleration = 4/9 m/s/s

(final velocity)^2 = (initial velocity)^2 + 2(acceleration)(distance)

(9)^2 = (0)^2 + (2)(4/9)d

d=91.125 m

____

However, I am almost certainly incorrect. I never took into account the incline of the angle (35 degrees), so I think my logic is flawed.
Yeah what happened to the a = g*sin35?
 
  • #7
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Ah, yes! This is exactly the piece of the puzzle I was missing. I had that written on my free body diagram; I haven't the faintest idea why I overlooked it!

Many thanks!
 

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