Finding work of a man running up stairs

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1. Aug 31, 2016

Pao44445

1. The problem statement, all variables and given/known data
During the Powerhouse lab, Jerome runs up the stairs, elevating his 102 kg body a vertical distance of 2.29 meters in a time of 1.32 seconds at a constant speed.

a. Determine the work done by Jerome in climbing the stair case.
b. Determine the power generated by Jerome.
(credit : http://www.physicsclassroom.com/ )
2. Relevant equations
W=Fdcosθ
P=W/t

3. The attempt at a solution
well, I got the right answers but I don't get it why I can directly use 102 kg for F=ma in W=Fdcosθ equation to find work and the problem doesn't give me any theta or I miss some concept of work?

2. Aug 31, 2016

haruspex

In that equation, what does theta represent in terms of the vectors $\vec F$ and $\vec d$?

3. Aug 31, 2016

Simon Bridge

The equation is $W=\vec s \cdot \vec F$ for the work done applying force F.
Notice that Jerome runs at constant velocity? Therefore the net force on him is zero.
Therefore your force diagram is incorrect.

4. Aug 31, 2016

Pao44445

An angle between Force and distance?

5. Aug 31, 2016

jbriggs444

With post #3 in mind, what is the magnitude and direction of the force?
What is the magnitude and direction of the distance?

6. Aug 31, 2016

Pao44445

I came back reading definition of work again
"Work results when a force acts upon an object to cause a displacement"
F that I must use in this equation is the force that the stairs act on him? (N) and distance has no direction since it is scalar quantity. I am confusing :/ Why I can use 2.29 m for d which it is the height of stair not the displacement of the man(Hypotenuse). I can use energy to solve this but this lesson is about work and I don't want to cheat too

7. Aug 31, 2016

haruspex

Yes, and in what direction is that?
Right, but the definition you quote says displacement, which is a vector.
It is fine to use the displacement along the slope of the stairway,provided you use the right vector for the force and take into account the angle between the two.
Now, you may object that you are not told the slope nor the distance up it of the displacement, but if you just create variables for those and apply a bit of trigonometry you should find they disappear from the equation.