Calculating the Vertical Height of Stairs Climbed by a 65.0 kg Student

[Brittykitty]

A 65.0 kg student climbs a set of stairs in 25.0 s with a power output of 95 W. What is the vertical height of the stairs?

I came up with the formula:

W=P x (delta t)
W=95w(25.0s)
2375

Im not sure how this comes into play:
W = mg (delta h)


Im really confused :|

 

[collinsmark]

You're on the right track.

Work is equal to force times distance. mg is the force. delta h is the distance.

[Edit: Technically, for a constant force, W = \vec F \cdot \vec s where the dot is the vector dot product using the total distance s. But here (in this problem) gravity points straight down. The component of s that is parallel to the force is delta h, the height of the stairs!]

 

[PhanthomJay]

A 65.0 kg student climbs a set of stairs in 25.0 s with a power output of 95 W. What is the vertical height of the stairs?

I came up with the formula:

W=P x (delta t)
W=95w(25.0s)
2375joules[/color]

the units for work are Newton-meters, or Joules

Im not sure how this comes into play:
W = mg (delta h)

That's right, this is the work done the climber against gravity, at assumed constant velocity.

Im really confused :|

Why? Solve for delta h.

Edit: confirming collinsmark reply.

 

[Brittykitty]

I'm not quite sure that I get it :(

So when I find the force that will be my height? This is the most confusing question to me lol

 

[Brittykitty]

How do I find delta h?

 

[collinsmark]

I'm not quite sure that I get it :(

So when I find the force that will be my height? This is the most confusing question to me lol

You had it in your original post. Earlier (in my previous post) I was simply confirming that your approach was correct.

You had already calculated,

W = P\Delta t

such that W = 2375 J. But you also worked out that

W = mg \Delta h,

therefore,

\mbox{2375 J} = mg \Delta h.

Solve for \Delta h, where \Delta h is the height of the stairs.

[I didn't mean to confuse you with the vector dot product stuff. All I was trying to point out was that it's not necessary to know the 3-dimensional length of the stairs. Since gravity is the only external force involved (well, besides normal forces and such) (and I'm also assuming that the student is ascending the stairs at a constant velocity [i.e. not accelerating] for simplicity), then gravity is the only force the student is "fighting." That means the height of the stairs (and not its overall length) is the only distance that matters.]
[Edit: Oh, and I'm also ignoring air resistance and the like. ]

 

[Brittykitty]

Thank you :)
so since
2375 J =mg(h) do I need to rearrange it so its h= ?

 

[Brittykitty]

m=65
g=(9.81)
(h)=?
J= 2375

I sort of it get it but still not sure.. terrible at physics

 

[Brittykitty]

h=(2375J)(9.81) ?

 

[cepheid]

h=(2375J)(9.81) ?

No, if you do anything to an equation, you have to do it to both sides of the equation (if you don't, then the two sides won't be equal anymore). For the equation:

W = mgh

...in order to leave just h on the right hand side, you need to divide the right hand side by mg. Therefore, you need to divide the left hand side by mg as well.

 

[Brittykitty]

2375 J =mgh

(65)(9.81)=637.65
2375J/637.65
=3.72?

 

[collinsmark]

2375 J =mgh

(65)(9.81)=637.65
2375J/637.65
=3.72?

Yeh! Your 3.72 m result looks good to me (but don't forget your units).

 

[Brittykitty]

Thank you so much! :) :) :)