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

Answer:

v_o = 4.54 m/s  

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

Answer:

2.10m/s

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 = $\frac{1}{2}$ x m x v²

$K_{E}$₁ = $\frac{1}{2}$ 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 =  ($\frac{1}{2}$ x m x v²) - ($\frac{1}{2}$ 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 =  ($\frac{1}{2}$ x m x 0²) - ($\frac{1}{2}$ x m x u²)

F x s cos θ =  0 - ($\frac{1}{2}$ x m x u²)

F x s cos θ =   - ($\frac{1}{2}$ 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 =   - ($\frac{1}{2}$ x m x u²)

μk x m x g x s cos θ = -  ($\frac{1}{2}$ x m x u²)

Divide through by m;

μk x g x s cos θ = -  ($\frac{1}{2}$ 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 = -  ($\frac{1}{2}$ x u²)

0.020 x 10 x 11 x (-1) = -  ($\frac{1}{2}$ x u²)

-2.2 = -  $\frac{1}{2}$ x u²

u² = 4.4

u = $\sqrt{4.4}$

u = 2.10m/s

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