- #1
Clever_name
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Homework Statement
Hi all, I am currently working on a problem that i believe involves the work-kinetic energy theorem.
See the attachment and information below.
As shown in the figure, the geometry of the water slide used is 1m:
Slide length = 61,; Slope angle = 24; Ramp height = 3.66 m
Lubricant is used thus friction can be ignored. A man of 70kg slides from the start.
if the ramp angle is 30, find the magnitude and direction of the velocity of this man when he just flies away from the ram and the Horizontal distance that the man flies.
Homework Equations
W=(1/2)mvf^(2) - (1/2)mvi^(2)
The Attempt at a Solution
ok so first I find the force that is acting on the man as he starts sliding down the ramp. which comes out to be mgcos(24) = F
and i know the distance of the slide which is 200ft. And further we can say W=FD
so subbing values in and solving for vf I get 108.43 ft/s - this is the velocity when the man is at the the lowest point on the slide.
So then i find i repeat the process and find vf at the top of the ramp with an angle of 30. which gives me an answer of 102.08 ft/s.
There is no answer to this question so i have no idea if I'm right or not... any help would be greatly appreciated.
also to find the direction of this velocity can i assume at the instant he comes off the slide then the angle the slide makes to the horizontal is also the direction of his velocity?
Thank you again!