Respuesta :
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:
2.10m/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[tex]K_{E}[/tex]) of the body. i.e
W = Δ[tex]K_{E}[/tex] [Δ[tex]K_{E}[/tex] = [tex]K_{E}[/tex]₂ - [tex]K_{E}[/tex]₁]
=> W = [tex]K_{E}[/tex]₂ - [tex]K_{E}[/tex]₁ ----------------(i)
[[tex]K_{E}[/tex]₂ = final kinetic energy [tex]K_{E}[/tex]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;
[tex]K_{E}[/tex]₂ = [tex]\frac{1}{2}[/tex] x m x v²
[tex]K_{E}[/tex]₁ = [tex]\frac{1}{2}[/tex] 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 [tex]K_{E}[/tex]₂ , [tex]K_{E}[/tex]₁ and W into equation (i) as follows;
F x s cos θ = ([tex]\frac{1}{2}[/tex] x m x v²) - ([tex]\frac{1}{2}[/tex] 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 = ([tex]\frac{1}{2}[/tex] x m x 0²) - ([tex]\frac{1}{2}[/tex] x m x u²)
F x s cos θ = 0 - ([tex]\frac{1}{2}[/tex] x m x u²)
F x s cos θ = - ([tex]\frac{1}{2}[/tex] 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 ([tex]F_{R}[/tex]) in the horizontal direction
i.e F = [tex]F_{R}[/tex]
But we know that the frictional force [tex]F_{R}[/tex], is given by;
F = [tex]F_{R}[/tex] = 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 = [tex]F_{R}[/tex] = 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 = - ([tex]\frac{1}{2}[/tex] x m x u²)
μk x m x g x s cos θ = - ([tex]\frac{1}{2}[/tex] x m x u²)
Divide through by m;
μk x g x s cos θ = - ([tex]\frac{1}{2}[/tex] 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 = - ([tex]\frac{1}{2}[/tex] x u²)
0.020 x 10 x 11 x (-1) = - ([tex]\frac{1}{2}[/tex] x u²)
-2.2 = - [tex]\frac{1}{2}[/tex] x u²
u² = 4.4
u = [tex]\sqrt{4.4}[/tex]
u = 2.10m/s
Therefore, the speed of the skier at the start of the slide is 2.10m/s
