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Sliding on a water slide

  • Thread starter vxr
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  • #1
vxr
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Homework Statement:

A person of mass ##m = 75## kg slides a distance ##d = 5## m on a straight water slide, dropping through a vertical height ##h = 25## m. Determine the mechanical work done by gravity on the person? What is the height h if the mechanical work done by the gravity is ##W = 2010## J?

Relevant Equations:

Probably some of these:
##E = mgh + \frac{1}{2} mv^2##
##W = \Delta \frac{1}{2} mv^2##
I don't really understand if the initial horizontal velocity is 0? Or do I assume it's some constant? Putting aside vertical velocity.

Also how should the "mechanical work done by gravity" be calculated? Is it just ##W = \frac{1}{2}mv^2_{final} - \frac{1}{2}mv^2_{initial}##
 
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Answers and Replies

  • #2
kuruman
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The mechanical work done by gravity is ##m\vec g \cdot \vec d## where ##\vec d ## is the displacement vector.
 
  • #3
BvU
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What does ##d=5## m mean ? in the context of dropping through 25 m !
 
  • #4
vxr
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I am really unsure. I tried to draw it and I came up with this:
243963
 
  • #5
kuruman
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The statement of the problem makes one believe that the slide is an inclined plane of 5 m hypotenuse and vertical side 25 m. This is impossible unless the vertical side is something else, 2.5 m perhaps?
 
  • #6
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Perhaps it should have said
A person of mass m=75 kg slides a horizontal distance d=5 m on a straight water slide, while dropping through a vertical height h=25 m.
?? That would mean the hypotenuse is sqrt(5^2 + 25^2) m.
 
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  • #7
haruspex
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Perhaps it should have said ?? That would mean the hypotenuse is sqrt(5^2 + 25^2) m.
That would make it far taller and steeper than any I've seen.
More likely the two numbers are swapped over: 5m vertical, 25m hypotenuse.
 

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