Is My Calculation of the Skier's Velocity at the Bottom of the Slope Correct?

[zomgwtf]

Kind of embarassing since I'm studying the sciences but someone came to me with help on this problem so I said I would do it and then try to help them out but I'm not sure if I did it correct 100% lol. yuuuuuuup.

Homework Statement

A 70kg skier slides down a 30-m slope that makes an angle of 28^\circ(how do you do the degree symbol lol) with the horizontal. Assume that \muk=0.2. What is the velocity of the skier at the bottom of the slope.

Homework Equations

gpe=ek+|Ff|

The Attempt at a Solution

So I rearranged the equation and got \sqrt{2gh-|Ff|/m}=v
where h=30sin28 and Ff=686cos28(0.2) (I like to use exact values instead of writing out long decimals)
So my final answer is around 16.6m/s is this correct?

Thanksss

 

[cepheid]

I get a different answer. There is something that jumps out right away as being obviously wrong with what you did:

Homework Equations

gpe=ek+|Ff|

It looks like you're trying to express conservation of energy, but there is an inconsistency. Everything in your equation has dimensions of energy except for the |Ff| term. From the expression you wrote for Ff in your post, I am inferring that you intended for Ff to be the frictional force (since you computed it by multiplying the normal force by the coeff. of friction). Remember that all of your equations must be dimensionally consistent. A statement like "energy = force" is nonsense.

Here's how I went about doing it. I want to apologize in advance for the ugliness of the equations. I didn't have an actual scratchpad on hand, so I used notepad on my computer to work out the problem (*before* reading your solution -- so that it would be an independent result). To save time, I didn't bother making these notepad equations look nicer. Also, I used kind of a weird practice from classical mechanics of calling the potential energy "U" and the kinetic energy "T." Anyway:

Givens:

m = 70 kg
d = 30 m
i = 28 deg. ("i" is for "inclination" angle)
u_k = 0.2

Solution:

E_before = E_after (cons. of energy)

U_g = T + E_lost (setting U_g = 0 at bottom of ramp for convenience)

Here E_lost is the mechanical energy lost due to friction (dissipated as heat). But the energy lost is equal to the negative of the work done by friction, so that if friction does negative work, a positive amount of energy is *lost*. Hence:

U_g = T - W_fric

T = U_g + W_fric [1]

Work out each of the terms in [1] separately:

U_g = mg*d*sin(i)
T = (1/2)m*v^2
W_fric = F_fric*d = -u_k*N*d (negative since this is actually a dot product and the force is opposite to the displacement)
N = mg*cos(i) (N is the normal force)

Now plug these four above things into [1]:

(1/2)mv^2 = mg*d*sin(i) + (-u_k*mg*cos(i)*d)

Factoring out the common variables on the right hand side:

(1/2)mv^2 = mg*d*[sin(i) - u_k*cos(i)]

The m's cancel:

v^2 = 2*g*d*[sin(i) - u_k*cos(i)]

v = {2*g*d*[sin(i) - u_k*cos(i)]}^(1/2)

EDIT: For the record, I get v = 13.13 m/s

 

[tiny-tim]

hi zomgwtf!

(have a degree: ° and a mu: µ and a square-root: √ and try using the X² and X₂ icons just above the Reply box )

your https://www.physicsforums.com/library.php?do=view_item&itemid=75" has no distance in it (and you need a factor 2 there)

 

[zomgwtf]

Hey cepheid thanks for the detailed response. I saw my mistake afterwards and I too got 13.13m/s.

Feels weird going back to grade 12 physics questions.