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Physics question

N

neonerd

Diamond Member
This is an optional problem because we didn't cover it ye, but i'd like to know how to do it...please show your steps 🙂

A ski starts from rest and slides down a 20 degree incline 100m long. If the coefficient of friction is 0.090, what is the ski's speed at the base of the incline?
 
f1 + f2 + f3 + f4 + ... = m*a. I hope you know this, the sum of forces in one direction is the mass times the acceleration of the body in that direction.

For gravity, f1, the force is m*g*sin(20°). Accleration from gravity is g. Use the proper sign of course to get gravity pulling in the proper direction. The angle is there to adjust for how much gravity is pulling parallel to the incline. Check: when the angle is 0°, the ground is level, and gravity should have no acceleration horizontally, thus f1 = m*g*sin(0°) = 0. Double check: when the angle is 90°, gravity should have its full effect as the object is in free fall, thus f1 = m*g*sin(90°) = m*g. Yep, in both cases the angle works as it should.

For friction, f2, the force is 0.090*m*g*cos(20°). The angle is there to adjust for how much gravity is pushing directly perpendicular to the incline. Check: When the ground is horzontal (0°), the friction should have its full effect: f2 = 0.090*m*g*cos(0°) = 0.090*m*g. Check: When the angle is (90°), there should be minimal contact and the body should fall without friction being a problem (assuming it isn't glued on in some form, but then the coefficient of friction would be infinity), thus f2 = 0.090*m*g*cos(90°) = 0. Yep the angle works as it should.

Solve for the acceleration.

Use a= dv/dt and v = dx/dt to solve for the velocity at the end.
 
Originally posted by: neonerd
would mass matter in this problem?
Click to expand...

You certainly aren't willing to do any work are you?

From my steps above:

f1 + f2 + f3 + f4 + ... = m*g*sin(20°) + 0.090*m*g*cos(20°) = m*a.

Solve for acceleration by dividing mass out of the equation above:

g*sin(20°) + 0.090*g*cos(20°) = a.

See how mass doesn't matter? It really wasn't any work at all to answer it yourself.
 
Originally posted by: dullard
Originally posted by: neonerd
would mass matter in this problem?
Click to expand...

You certainly aren't willing to do any work are you?

From my steps above:

f1 + f2 + f3 + f4 + ... = m*g*sin(20°) + 0.090*m*g*cos(20°) = m*a.

Solve for acceleration by dividing mass out of the equation above:

g*sin(20°) + 0.090*g*cos(20°) = a.

See how mass doesn't matter? It really wasn't any work at all to answer it yourself.
Click to expand...

i did...just wasn't able ti update this thread from physics class 😛

it was posted during spanish 😉

thanks for your help 🙂
 
Originally posted by: Triumph
Originally posted by: IHateMyJob2004
Originally posted by: Howard
Originally posted by: neonerd
would mass matter in this problem?
Click to expand...
No.
Click to expand...

real world, yes. Theory, no.
Click to expand...

You mean the laws of physics only apply to homework problems and not to the real world? :Q
Click to expand...


No, just that the way that the problem was described (assumtion of constant co-efficient of friction).

Note that for most purposes (ie the range of human body weights) the assumption pretty much holds out, it is only when the mass gets large enough to change the co-efficient of friction (or conversly light enough that air resistance becomes significant), but the mathemetics involved is extremelty complex. The problem is about learning the basics first.
 
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