Question

Find the coefficient of kinetic friction μk. express your answer in terms of some or all of the variables d1, d2, and θ.

Answer 1

internet tension = mass * acceleration internet tension = 23 – Friction tension = 14 * acceleration Friction tension = µ * 14 * 9.8 = µ * 137.2 23 – µ * 137.2 = 14 * acceleration Distance = undemanding speed * time undemanding speed = ½ * (preliminary speed + very final speed) Distance = ½ * (preliminary speed + very final speed) * time Distance = 8.a million m, preliminary speed = 0 m/s, very final speed = a million.8 m/s 8.a million = ½ * (0 + a million.8) * t Time = 8.a million ÷ 0.9 = 9 seconds Acceleration = (very final speed – preliminary speed) ÷ time Acceleration = (a million.8 – 0) ÷ 9 = 0.2 m/s^2 23 – µ * 137.2 = 14 * 0.2 resolve for µ

Answer 2

μ_k = (d_1)*sin(Q) / [(cos(Q))*(d_1) + (d_2)]

Step-by-step explanation:

On the slope:

- The force of gravity has magnitude "mg" and points points straight down the slope and is given by the expression:

F_g,down = m*g*sin(Q)

- The work done by gravity W_g is the dot product of F_g and the distance traveled d_1, shown:

W_g = F_g . d_1

W_g = m*g*sin(Q). d_1

- The normal Force F_n acts perpendicular to the slope and the gravity components also acts in the opposite to its direction, apply Equilibrium condition in the axis perpendicular to the slope since, their is no acceleration in that direction:

F_perp = F_n - F_g,up = 0

F_n = F_g,up = m*g*cos(Q)

- Since there is no work done in the direction perpendicular to slope its zero.

- Frictional force F_f acts parallel to the slope opposite in direction of motion, hence, directed up the slope is dependent on coefficient of friction u_k and normal contact force as follows:

F_f = u_k . F_n

F_f = u_k (m*g*cos(Q))

- The work done by friction W_f is the dot product of F_f and the distance traveled d_1, shown:

W_f = F_f . d_1

W_f = u_k*m*g*cos(Q). d_1

- So in summary, the total work done on the sled during the downhill phase is:

W_downhill = (m*g)*(d_1)*sin(Q) – (m*g)*(cos(Q))*(μ_k)*(d_1)

The horizontal terrain:

- Since, the Force of gravity (F_g) and normal force (F_n) both act perpendicular to the direction of motion, they do zero work.

- However, friction force (F_f) acts parallel to the direction of motion but opposite in direction, the work done by friction is:

W_f = F_f*d_2

W_f = u_k*F_n * d_2

W_f = - u_k * m*g*d_2

- So the total work done during the horizontal phase is:

W_hor = W_f = - u_k * m*g*d_2

Total Work done:

- So that means the total work done on the sled overall is:

W = W_downhill + W_hor

= (m*g)*(d_1)*sin(θ) – (m*g)*(cos(Q))*(μ_k)*(d_1) – (m*g)*(μ_k)*(d_2)

- The total work done is zero because the Kinetic energy at the start and end of the journey is zero.

(m*g)*(d_1)*sin(θ) – (m*g)*(cos(Q))*(μ_k)*(d_1) – (m*g)*(μ_k)*(d_2) = 0

- Simplify:

(d_1)*sin(Q) = (cos(Q))*(μ_k)*(d_1) + (μ_k)*(d_2)

(d_1)*sin(Q) = [(cos(Q))*(d_1) + (d_2)]*(μ_k)

- And then:

μ_k = (d_1)*sin(Q) / [(cos(Q))*(d_1) + (d_2)]