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
