Calculating Power to Drag 100kg Log Down 20° Hillside at 0.50 m/s

[NoMeGusta]

What power must a man expend on a 100-kg log that he is dragging down a hillside at a speed of 0.50 m/s ? The hillside makes an angle of 20 degrees with the horizontal and the coefficient of friction is 0.9

m = 100kg
v = 0.50 m/s
$$\theta$$ = 20 degrees
$$\mu$$ = 0.9

From here I thought that $$W_{man} + W_{f} = \Delta U$$

So, $$W_{f} = - \mu mgL \cos{\theta}$$ (that I understood), from here the books states that $$\Delta U = 0 - mgL \sin{\theta}$$ Where did the book get this? What does the 0 represent? How do I find the Power?

The book then says that $$P_{man} = \mu mgv \cos{\theta} - mgv \sin{\theta} = 247J$$

Can someone walk me thru this?

 

[Chi Meson]

U is the proper variable symbol for potential energy, gravitational potential energy in this case.

The $$mgL \sin{\theta}$$ is another way of say "mgh," but on this slope (think triangle), h is the opposide side where L is the hypotenuse. mgh is the change in potential energy. The work done by the guy has to do two things: change the potential energy of the log, and overcome friction.

Conceptual rant:
Depending of conventions and definitions, sometimes the work done to overcome friction is not called "work" but "the mechanical equivalent of heat." That last bit is a mouthful, and I haven't seen it too often in textbooks lately, so I guess it's OK to call it work; the only problem is that it is not a transfer of mechanical energy if it turns to thermal energy, it should be called heat. Alas this distinction is low on the scale of hair-splitting importance.[/concpetual rant]

 

[NoMeGusta]

What instrument do you play hemidemisimiquavers? I thought 16th notes were fast enough... geez, I can only imagine 64ths

 

[Chi Meson]

What instrument do you play hemidemisimiquavers? I thought 16th notes were fast enough... geez, I can only imagine 64ths

I can handle short trills on the piano that are "hemis." As long as its with fingers 2 & 3 (the thumb is #1 on the piano)