- #1

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## Homework Statement

What minimum speed does a 136.4 g puck need to make it to the top of a 3.31-m-long, 24° frictionless ramp?

## Homework Equations

K=mV^2/2 U=mgY

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- Thread starter talaroue
- Start date

- #1

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What minimum speed does a 136.4 g puck need to make it to the top of a 3.31-m-long, 24° frictionless ramp?

K=mV^2/2 U=mgY

- #2

rock.freak667

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the 'y' in your equation is the vertical distance, not horizontal. So the cos24 should be sin24

- #3

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cos angle = b/c

f * s + potential energy = mv^2 / 2

edit: mistake if you want height then its sin.

- #4

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gh = v^2/2

9.82 * 1.35m = v^2 / 2

26.5 =v^2

v = 5.15m/s

but it think this is totally wrong.

- #5

Cyosis

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- #6

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ahhhhh. Thanks guys i see where i went wrong you guys are awesome!!

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