Answer:
μk = 0.124
Explanation:
Known data
m=5.0 kg : mass of the sled
T= 10 N : force with which the boy pulls the rope
θ =60.0° :angle of the rope with respect to the horizontal direction
g = 9.8 m/s² : acceleration due to gravity
Newton's second law to the sled :
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
Forces acting on the sled
W: Weight of the sled : In vertical and downward direction
N : Normal force : In vertical and upwards direction
f : Friction force: parallel to the movement of the sled and in the opposite direction to the movement
T:Rope tension : forming angle 60.0° of of the rope with respect to the horizontal direction
Calculated of the W of the sled
W= m*g
W= 5.0 kg* 9.8 m/s² = 49 N
x-y components of the tension of the rope T
Tx= 10*cos60°= 5 N
Ty= 10*sin60° = 8.66 N
Calculated of the N
We apply the formula (1)
∑Fy = m*ay ay = 0
N+Ty -W = 0
N = 49 N - 8.66 N
N = 40.34 N
Calculated of the f
f = μk* N
f = μk* 40.34 Equation (1)
We apply the formula (1) to calculated f
∑Fx = m*ax the sled moves with constant velocity, then ax=0
∑Fx = 0
Tx-f = 0
5 - f = 0
f = 5N
We replace f in the equation (1)
5 = μk* 40.34
μk = 5 / 40.34
μk = 0.124
The coefficient of friction between the sled and the ice is 0.124.
The given parameters;
The resultant normal force of the rope is calculated as;
[tex]F_n + Tsin(\theta) - W=0[/tex]
[tex]F_n = mg -Tsin(\theta)\\\\F_n = (5\times 9.8) - 10sin(60)\\\\F_n = 40.34 \ N[/tex]
The resultant horizontal force on the rope is calculated as;
[tex]Tcos(\theta) -F_k = 0\\\\F_k = Tcos(\theta) \\\\F_k = 10 \times cos (60)\\\\F_k = 5 \ N[/tex]
The coefficient of friction between the sled and the ice is calculated as;
[tex]F_k =\mu_k F_n\\\\\mu_k = \frac{F_k}{F_n} \\\\\mu_k = \frac{5}{40.34} \\\\\mu_k = 0.124[/tex]
Thus, the coefficient of friction between the sled and the ice is 0.124.
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