Respuesta :
Answer:
a) μ = 0.0398
b) μ = 0.1595
Explanation:
a) d = 184 m
Vi = 12 m/s
vf = 0 m/s
m = 61 Kg
μ = ?
We use the equation
W = Kf - Ki (I)
then we have
W = F(friction)*d = - μ*N*d = - μ*m*g*d (II)
Kf = m*(vf)²/2 = m*(0)²/2 = 0 J
If (I) = (II)
0 - m*(vi)²/2 = - μ*m*g*d ⇒ μ = (vi)² / (2*g*d)
⇒ μ = (12 m/s)² / (2*9.81 m/s²*184 m) = 0.0398
b) d = 184 m
Vi = 24 m/s
vf = 0 m/s
m = 75 Kg
μ = ?
We use the equation
W = Kf - Ki (I)
then we have
W = F(friction)*d = - μ*N*d = - μ*m*g*d (II)
Kf = m*(vf)²/2 = m*(0)²/2 = 0 J
If (I) = (II)
0 - m*(vi)²/2 = - μ*m*g*d ⇒ μ = (vi)² / (2*g*d)
⇒ μ = (24 m/s)² / (2*9.81 m/s²*184 m) = 0.1595
The mass of the skier is not important in the analysis.
A) The coefficient of kinetic friction between the skis and snow is; μ_k = 0.04
B) The coefficient of kinetic friction between the skis and snow when mass is 75 kg and initial speed is twice that in A above is; μ_k = 0.1597
We are given;
Mass of skier; m = 61 Kg
Distance; d = 184 m
Initial speed; u = 12 m/s
Final speed; v = 0 m/s
- A) The work energy theorem formula is given as;
W_net = Final kinetic energy - initial kinetic energy
Thus;
W_net = K_f - K_i
Also; Net work from friction is;
W_net = μ_k•mgd
Where μ_k is coefficient of kinetic friction
Thus;
μmgd = K_f - K_i
K_f = ½mv²
K_f = 0 m/s because final velocity is 0 m/s
K_i = ½mu²
Thus;
μ_k•mgd = ½mu²
m will cancel out to give;
μ_k•gd = ½u²
μ_k = ½u²/gd
Plugging in the relevant values gives;
μ_k = ½ × 12²/(9.8 × 184)
μ_k = 0.04
- B) Now, the parameters have changed to;
m = 75 kg
u = 2(12) = 24 m/s
Thus;
μ_k = ½ × 24²/(9.8 × 184)
μ_k = 0.1597
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