Question

A cross-country skier slides horizontally along the snow and comes to rest after sliding a distance of 11 m. If the coefficient of kinetic friction between her skis and the snow is μk = 0.020, determine her speed at the start of the slide.

Answer 1

v_o = 4.54 m/s

Step-by-step explanation:

Knowns

From equation, the work done on an object by a constant force F is given by:

W = (F cos Ф)S (1)

Where S is the displacement and Ф is the angle between the force and the displacement.

From equation, the kinetic energy of an object of mass m moving with velocity v is given by:

K.E=1/2m*v^2 (2)

From The work- energy theorem , the net work done W on an object equals the difference between the initial and the find kinetic energy of that object:

W = K.E_f-K.E_o (3)

Given

The displacement that the sled undergoes before coming to rest is s = 11.0 m and the coefficient of the kinetic friction between the sled and the snow is μ_k = 0.020

Calculations

We know that the kinetic friction force is given by:

f_k=μ_k*N

And we can get the normal force N by applying Newton's second law to the sled along the vertical direction, where there is no acceleration along this direction, so we get:

∑F_y=N-mg

N=mg

Thus, the kinetic friction force is:

f_k = μ_k*N

Since the friction force is always acting in the opposite direction to the motion, the angle between the force and the displacement is Ф = 180°.

Now, we substitute f_k and Ф into equation (1), so we get the work done by the friction force:

W_f=(f_k*cos(180) s

=-μ_k*mg*s

Since the sled eventually comes to rest, K.E_f= 0 So, from equation (3), the net work done on the sled is:

W= -K.E_o

Since the kinetic friction force is the only force acting on the sled, so the net work on the sled is that of the kinetic friction force

W_f= -K.E_o

From equation (2), the work done by the friction force in terms of the initial speed is:

W_f=-1/2m*v^2

Now, we substitute for W_f= -μ_k*mg*s, and solving for v_o so we get:

-μ_k*mg*s = -1/2m*v^2

v_o = √ 2μ_kg*s

Finally, we plug our values for s and μ_k, so we get:

v_o = √2 x (0.020) x (9.8 m/s^2) x (11.0 m) = 4.54 m/s

v_o = 4.54 m/s

Answer 2

2.10m/s

Step-by-step explanation:

Here, we use the work-energy principle that states that the work done (W) on a body is equal to the change in kinetic energy (Δ
`K_(E)`) of the body. i.e

W = Δ
`K_(E)` [Δ
`K_(E)` =
`K_(E)`₂ -
`K_(E)`₁]

=> W =
`K_(E)`₂ -
`K_(E)`₁ ----------------(i)

[
`K_(E)`₂ = final kinetic energy
`K_(E)`₁ = initial kinetic energy]

But;

W = F x s cos θ

Where;

F = net force acting on the body

s = displacement of the body due to the force

θ = angle between the force and the displacement.

Also;

`K_(E)`₂ =
`(1)/(2)` x m x v²

`K_(E)`₁ =
`(1)/(2)` x m x u²

Where;

m = mass of the body

v = final velocity of the body

u = initial velocity of the body

Substitute the values of
`K_(E)`₂ ,
`K_(E)`₁ and W into equation (i) as follows;

F x s cos θ = (
`(1)/(2)` x m x v²) - (
`(1)/(2)` x m x u²) -----------------(ii)

From the question;

i. The skier comes to a rest, this implies that the final velocity (v) of the body(skier) is 0.

Therefore substitute v = 0 into equation (ii) to get;

F x s cos θ = (
`(1)/(2)` x m x 0²) - (
`(1)/(2)` x m x u²)

F x s cos θ = 0 - (
`(1)/(2)` x m x u²)

F x s cos θ = - (
`(1)/(2)` x m x u²) ---------------------(iii)

ii. Since there is no motion in the vertical direction, the net force (F) acting is the kinetic frictional force (
`F_(R)`) in the horizontal direction

i.e F =
`F_(R)`

But we know that the frictional force
`F_(R)`, is given by;

F =
`F_(R)` = μk x N

Where;

μk = coefficient of static friction

N = Normal reaction which is equal to the weight (m x g) of the skier [since there is no motion in the vertical]

=> F =
`F_(R)` = μk x m x g [m = mass of the skier and g = acceleration due to gravity]

iii. Also, since the only force acting is the frictional force acting to oppose motion, the angle θ between the force and the displacement is 180°

iv. Now substitute all of these values into equation (iii) as follows;

F x s cos θ = - (
`(1)/(2)` x m x u²)

μk x m x g x s cos θ = - (
`(1)/(2)` x m x u²)

Divide through by m;

μk x g x s cos θ = - (
`(1)/(2)` x u²) ----------------(iv)

From the question;

s = 11m

μk = 0.020

Take g = 10m/s²

θ = 180°

Substitute these values into equation (iv) and solve for u;

0.020 x 10 x 11 cos 180 = - (
`(1)/(2)` x u²)

0.020 x 10 x 11 x (-1) = - (
`(1)/(2)` x u²)

-2.2 = -
`(1)/(2)` x u²

u² = 4.4

u =
`√(4.4)`

u = 2.10m/s

Therefore, the speed of the skier at the start of the slide is 2.10m/s