- #1

pinsky

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## Homework Statement

The stairs (moving stairs like the ones in a mall) are moving with the vertical component of v

_{s}=1,5 m/s. What power should a man with a mass of m=80kg use to climb up to a height h=12m in t=5s using the stairs?

## Homework Equations

g=9.81

Book solution:

[tex] P=\frac {mgh_p}{t} [/tex]

[tex]h_p=h-v_st [/tex]

[tex]P=\frac {mg(h-v_s t)}{t} [/tex]

They calculated the height which the stairs take the man to, from which they got the height the man has to travel on his own. They used that height to calculate work.

## The Attempt at a Solution

The average speed the man should go (with the speed of the stairs) is:

[tex] v_{tot}=\frac {h}{t} [/tex]

from which i can get the speed which the man should be going on his own.

[tex]v_{man}=v_{tot}-v_{stairs} [/tex]

So the kinetic energy of the man traveling at speed v

_{man}is:

[tex]E_{k-man} = \frac {1}{2} m v_{man} [/tex]

That is also the work the man does since his kinetic energy was 0 at the bottom of the stairs.

So power should be:

[tex]P=\frac{E_{k-man}}{t} [/tex]

I don't get the same result as they do.

My other assumption was that:

[tex] E_p=E_{k-stairs}+E_{k-man} [/tex]

[tex] mgh=\frac{1}{2}m(v_{man}^2+v_{stairs}^2) [/tex]

From which I get that:

[tex] v_{man}=\sqrt {2gh-v_{stairs}^2 [/tex]

form which it turns out that the man's speed should be grater than the h/t.

Where am i wrong?

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