Calculating Work and Power in Basic Physics: A Case Study

[WillyTech]

Ok, here's the deal, this is a physics problem that might seem basic to you. But to me its a pain, I've been working on this problem for 2 days now, and my group members are of no assistance to me. They have different answers than i do which i didn't agree to.

The question
Jessica runs up a 5 meter high stair to get to class on time. Her mass is 55kg and it takes her 10 seconds to run up the stairs.
A. What is her force?
B. How much work does she do?
C. How much power does she produce?

Heres the work that i have already put into and i am not sure if I am wrong or not.

Force- since F=m*a (due to gravity) i took 55kg multiplied to 10m/s2.
Since gravity pulls at roughly 9.8m/s2 -> i round to 10m/s2
55kg*10m/s2 = 550kg m/s2
(And i said that kg m/s2 is=to (Newtons)
So my force is 550N

Work- since W=F*d I took my force 550N and multiplied to 5m.
N*m = Joules
550N*5m = 2750 J
So my work is 2750J

Power- since P=w/t I took my work 2750J and divide by my time 10sec.
work/time=watts
2750J/10s = 275watts
So my power is 275Watts

I am not sure that these are correct but its been killing me for 2 days just thinking about it. And i know you guys out there can help.

 

[berkeman]

Looks okay to me so far. Thanks for posting your work so far -- that makes it a lot more practical for us to help you. BTW, don't round 9.8 to 10.0 for these calcs. 9.8 is a good solid number to use. Keep up the good work.

 

[WillyTech]

Ok, ill keep the 9.8 in mind. Thanks

Willy

 

[Pengwuino]

Yes that is correct. Your work will be defined as:

W = \int_{x1}^{x2} {Fdx}

It is also defined as

W = \Delta U + \Delta KE

Since there is no change in kinetic energy overall, you work with the potential energy. If she is 55kg and is 5m up, you can determine the work done by her. You can then determine power by this. Since your initial x value is 0, a simple F*dx calculation is accurate.