How much work is done on the skier in the first 8.9 seconds of descent?

[The Merf]

ok guys, i got one that wants me to figure out how much work is done:
A skier of a mass 79.1 kg, starting from rest, slides down a slope at an angle of 38 degrees with the horizontal. The coefficient of kinetic friction, u, is 0.09. What is the net work (the net gain in kinetic energy) done on the skier in the first 8.9 s of decent?
now i don't want the answer, i just need help know what equations i need when.

I know it gives me m=79.1kg theta=38degrees coefficient of kinetic friction=.09 (not sure of unit) and t=8.9

how do i get how much work?

I have figured the mass in the verticle direction to be 48.69 (is this right?) by taking sin38=x/79.1 or 79.1sin38=x

 

[Doc Al]

Figure out the acceleration. What's the net force on the skier? Use Newton's 2nd law.

 

[The Merf]

ok since i got 48.69 to be the mass in the y direction, i take that times -9.8 as the acceleration due to gravity and get -477.16 as the force, i think i Newtons, but where does friction come in?

 

[The Merf]

I can complete this problem without friction, or even with friction, but in one dimention, but what confuses me is that usually gravity and friction work together to slow down an object, now gravity is working to speed it up and friction is trying to slow it down, so I'm not sure what equation to use

 

[Doc Al]

ok since i got 48.69 to be the mass in the y direction, i take that times -9.8 as the acceleration due to gravity and get -477.16 as the force, i think i Newtons, but where does friction come in?

Don't think in terms of mass in a particular direction--that doesn't make much sense. Think of the component of the weight (mg) parallel to the incline, which equals mgsinθ.

Friction is another force acting on the skier; it will equal μN, where N is the normal force. (How do you determine the normal force?)

 

[Doc Al]

now gravity is working to speed it up and friction is trying to slow it down, so I'm not sure what equation to use

You need the net force down the incline. Since gravity and friction act in opposite directions, they will have opposite signs.

 

[The Merf]

ok so N=mgcos(theta)
N=79.1x9.8xCos(38)
N=477.25 (i think it is a little different because of rounding earlier)


so friction=uN
friction=-0.09x477.25
friction=-42.95

so the net forces N+friction=477.25-42.95 or 434.30 so that is the net force

now K=mv^2
w=dK=(K-K_0)/2
so 1/2K (because he starts at rest so v_0=0 so K_0=0)
or (mv^2)/2
now m=79.1kg, v=axt or because a=F/m v=434.30/79.1x8.9 or 48.87m/s
so K=1/2(79.1)(48.87)^2
w=94,456.35j
(is this right and is joules the right unit?)

 

[Doc Al]

ok so N=mgcos(theta)
N=79.1x9.8xCos(38)
N=477.25 (i think it is a little different because of rounding earlier)

Redo this calculation. (Cosine, not sine!)

 

[The Merf]

oops... my bad, i did use sin i just accidentally wrote cos
so is it supposed to be cos?

 

[Doc Al]

oops... my bad, i did use sin i just accidentally wrote cos

You should have used cosine, as written.

 

[The Merf]

ok thanks

 

[The Merf]

ok so N=mgcos(theta)
N=79.1x9.8xCos(38)
N=610.85

so friction=uN
friction=-0.09x610.85
friction=-54.98

so the net forces N+friction=610.85-54.98 or 422.27 so that is the net force

now K=mv^2
w=dK=(K-K_0)/2
so 1/2K (because he starts at rest so v_0=0 so K_0=0)
or (mv^2)/2
now m=79.1kg, v=axt or because a=F/m v=422.27/79.1x8.9 or 47.51m/s
so K=1/2(79.1)(47.51)^2
w=89,272.26j
so joules is the right unit?

 

[Doc Al]

ok so N=mgcos(theta)
N=79.1x9.8xCos(38)
N=610.85

so friction=uN
friction=-0.09x610.85
friction=-54.98

OK.

so the net forces N+friction=610.85-54.98 or 422.27 so that is the net force

The net force is the component of gravity down the incline (not N) minus the friction force.