Answer:
v_o = 4.54 m/s  
Explanation:
<u>Knowns  </u>
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)
<u>Given </u>
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  
<u>Calculations</u> 
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