Dynamics: Skater on icy road

1. Definitions

Let's recall:

1. A friction force ƒ on a moving object, is equal to the kinetic friction coefficient μ_k times the normal force. If the normal force equates the weight mg, then:
N = mg and ƒ = μ_k N = μ_k m g.

2. The friction force is a non-conservative force. The work done by this force is always negative.

3. The work done by a conservative force (such as gravitational force) on a moving object does not depend on the path taken by this object. It depends only on the initial and final position of the object.

4. The work done by any force F on a moving object is equal to the scalar product of this force by the displacement l of the object:
W = (vector)F . (vector)l = F l cos θ.
θ is the angle between the two vectors F and l.
Therefore The force normal to the displacement does not work. The only forces that work are parallel to the displacement.

5. The kinetic energy theorem holds for the conservative forces and for the non-conservative forces. It is written as:
ΔKE = Σ W(F_ext)

Δ KE is the difference between the final kinetic energy and the initial kinetic energy of the object of mass m:
= (1/2) m v_f² - (1/2) m v_i²
Where
v_i and v_f are the initial and final velocities of the object respectively.
W(F_ext) is the work done by all the external forces that act on the object(such as gravitational force, normal force, friction force, ...)

6. The energy conservation principle holds for only for the conservative forces. It says: The total (or mechanical) energy E of a system, which is equal to the sum of the kinetic energy KE and the potential energy PE is conserved.
It is written as:
E = KE + PE = constant, or
KE_initial + PE_initial = KE_final + PE_final

1. Skater on icy road

1. Part CE

At the point C, the skater has an horizontal velocity v_C.

The height (CD) is written as
(CD) = (1/2) g t²

The horizontal range (DE) is written as (DE) = v_C t

Therefore v_C = (DE)/t

Squaring gives: v_C² = (DE)²/t²

Substituting "t" given by the first relationship, we get:

v_C² = g (DE)²/2(CD)    (1)

2. Part BC

Applying the Kinetic energy theorem from B to C yields:
(1/2) m [v_C² - v_B²] = (F_f) (x_C - x_B) = - (m g μ_k)(BC)

Therefore
(1/2) [v_C² - v_B²] = - (g μ_k)(BC)

Hence
v_B² = v_C² + (2g μ_k)(BC)    (2)

Remark:
We can use here the Kinematics formula:
v_c² - v_B² = 2 a (x_C - x_B)
Where "a" is the acceleration of the skater given by Newton's second law:
Σ F = m a.

We have: Σ F = Friction force = - ƒ ( minus sign because it is directed in the opposite side of the acceleration, which is the direction of the motion). Therefore
m a = - ƒ = - m g μ_k. So
a = - g μ_k.

Substituting "a" in the above Kinematic formula gives:
v_c² - v_B² = - 2 g μ_k (x_C - x_B)
That is:
v_c² + 2 g μ_k (BC) = v_B²,
Which is exactly the same found formula (2)


3. Part AB

Applying the Kinetic energy theorem from A to B, along the incline plane yields:



ΔKE = Σ Work(F_ext)

The F_ext are: mg, N, and ƒ

N does not work.

The net force that works is
m g sin 25 - ƒ = m g sin 25 - μ_k m g cos 25.

This net force does a work over the inclined distance "l". This work is:
[m g sin 25 - μ_k m g cos 25] l.

Therefore

(1/2) m[v_B² - 0 ] = [m g sin 25 - μ_k m g cos 25] l

with h/l = sin 25,
that is l = h/sin 25, we get:

[v_B² = 2 [ g sin 25 - μ_k g cos 25] (h/sin 25)
[v_B² = 2 g [sin 25 - μ_k cos 25] (h/sin 25)

or

v_B² = 2 g h [1 - μ_k cot 25]    (3)

Recall: cot θ = 1/tan θ = cos θ /sin θ

4. Rearranging the above results:

Using (2) gives:
v_C² + (2g μ_k)(BC) = 2 g h [1 - μ_k cot 25]

Using (1), gives:
g (DE)²/2(CD) = - (2g μ_k)(BC) + 2 g h [1 - μ_k cot 25]
[g (DE)²/2(CD) + (2g μ_k)(BC)] = 2 g h [1 - μ_k cot 25]
h = [g (DE)²/2(CD) + (2g μ_k)(BC)] / 2 g [1 - μ_k cot 25]

We have:
CD = DE = 1.50

Then

h = [(DE)/4 + (μ_k)(BC)] / [1 - μ_k cot 25]

With
BC = 10 m
μ_k = 0.402

h = [(1.50)/4 + (0.402)(10)] / [1 - 0.402 cotg 25]
= (0.375 + 4.02)/(1 - 0.862) = 31.85 m
h = 31.85 m

h = 31.85 meters