Time vs distance Question using Force

[wilson_chem90]

Homework Statement

A sight seen on many bunny hills across Ontario is young skiers pushing ski poles and gliding down a slope until they come to rest. Observing from a distance, you note a young person (approximately 25 kg) pushing off with the ski poles to give her an initial velocity of 3.5 m/s. If the inclination of the hill is 5.0 degrees and the coefficient of kinetic friction for the skis on dry snow is 0.20 calculate:

a) the time taken for the skier to come to a stop
b) the distance traveled down the hill

Homework Equations

Ff = ukFn
Fw = mg
F = m (vf-vi/t)
F = m (vf^2 - vi^2 / 2d)

The Attempt at a Solution

a) Fk = ukmg (Fn)
= .20(25)(9.8m/s^2)
= 49 N
F = m (vf - vi / t)
t = m (vf - vi / F)
= 25kg (0 m/s - 3.5 m/s / 49 N)
t = 1.79 s

b) F = m (vf^2 - vi^2 / 2d)
d = m(vf^2 - vi^2 / 2F)
= 25kg ((0m/s)^2 - (3.5m/s)^2 / 2(49N))
d = 12.5 m

 

[wilson_chem90]

sorry i calculated the distance incorrectly the distance is actually 3.1 m, which does not make much sense, so can someone please help me?

 

[PhanthomJay]

You have not correctly identified all the forces acting on the skier. Your relevant equations are fine, noting that F should be F_net, but the normal force (perpendicular to the plane) is not mg, and more than just the friction force acts along and parallel to the plane. You must take into account the 5 degree slope. Choose the x-axis along the slope, and the y-axis perpendicular to the slope, break up the weight force into is x and y components, and apply Newton's laws in each direction, noting that in the y direction, there is no acceleration.

 

[wilson_chem90]

alright so i tried that and i get this for Fnet

Fn - Fgcos5 degrees = 0
Fn = Fg cos5
= mg cos5
= 25kg(9.8m/s^2) cos5
= 244.1N

Ff - Fgsin5 = 0
Ff = mg sin5 degrees
= 25kg(9.8m/s^2) sin5 degrees
= 21.3N

F net = Ff + Fn
= 244.1N + 21.3N
= 265.4N

 

[PhanthomJay]

alright so i tried that and i get this for Fnet

Fn - Fgcos5 degrees = 0
Fn = Fg cos5
= mg cos5
= 25kg(9.8m/s^2) cos5
= 244.1N

good.

Ff - Fgsin5 = 0

The skier is decelerating; the net force up and parallel to the plane cannot be 0.

Ff = mg sin5 degrees
= 25kg(9.8m/s^2) sin5 degrees
= 21.3N

correct this; Ff = u_k(F_n)

F net = Ff + Fn
= 244.1N + 21.3N
= 265.4N

No, you have to look at F_net in the x direction only when calculating the acceleration in the x direction; What is the net force in the x direction, then proceed per your original post to find the acceleration, time, and distance.