150n sled is pulled by 28° slope with constant speed by 100n force. What is the cofficent of kinatic friction slead and the slope.
normal force on ground = m g cos 28 = 150 * 0.883 = 132.4 Newtons
so 133.4 * mu = friction force back down slope
component of weight down slope = m g sin 28 = 150*0.469 = 70.4 Newtons
so
133.4 * mu + 70.4 = 100
133.4 mu = 29.6
mu = 0.222
To find the coefficient of kinetic friction between the sled and the slope, we need to use the information given:
Force of gravity acting on the sled (Fg) = 150 N
Force applied to pull the sled (Fapplied) = 100 N
Angle of the slope (θ) = 28°
First, let's calculate the component of the force of gravity acting down the slope (Fg_parallel) and the component perpendicular to the slope (Fg_perpendicular).
Fg_parallel = Fg * sin(θ)
Fg_parallel = 150 N * sin(28°)
Fg_parallel ≈ 70.19 N
Fg_perpendicular = Fg * cos(θ)
Fg_perpendicular = 150 N * cos(28°)
Fg_perpendicular ≈ 132.68 N
Since the sled is pulled up the slope with a constant speed, the force of kinetic friction acting down the slope (Ff) must balance the component of the force applied up the slope (Fapplied_parallel).
Ff = Fapplied_parallel
Fapplied_parallel = Fapplied * cos(θ)
Fapplied_parallel = 100 N * cos(28°)
Fapplied_parallel ≈ 89.80 N
Therefore, the coefficient of kinetic friction (μk) can be calculated by dividing the force of kinetic friction by the perpendicular force.
μk = Ff / Fg_perpendicular
μk = 89.80 N / 132.68 N
μk ≈ 0.676
So, the coefficient of kinetic friction between the sled and the slope is approximately 0.676.
To find the coefficient of kinetic friction between the sled and the slope, we can use the formula:
μ = (Fk - Fpar)/Fn
Where:
μ is the coefficient of kinetic friction,
Fk is the force of kinetic friction,
Fpar is the component of the force parallel to the slope,
Fn is the normal force.
First, let's find the component of the force parallel to the slope (Fpar). We can calculate it using trigonometry:
Fpar = F * sin(θ)
= 100 N * sin(28°)
Fpar ≈ 47.587 N
Next, let's find the normal force (Fn). The normal force is the force exerted perpendicularly by the surface, which is equal in magnitude and opposite in direction to the component of the sled's weight that acts perpendicular to the slope. Since the sled is on a slope, part of its weight will be acting parallel to the slope:
Fperpendicular = m * g * cos(θ)
= 150 N * 9.8 m/s^2 * cos(28°)
Fperpendicular ≈ 400.734 N
Now, the normal force (Fn) is the opposite of the component acting parallel to the slope:
Fn = -Fperpendicular
≈ -400.734 N
Finally, we can substitute these values into the equation for the coefficient of kinetic friction:
μ = (Fk - Fpar) / Fn
= (Fk - 47.587 N) / -400.734 N
Since the sled is pulled with a constant speed, we know that the force of kinetic friction is equal to the applied force. Therefore:
100 N - 47.587 N = Fk
Fk ≈ 52.413 N
Substituting this value back into the equation, we get:
μ = (52.413 N - 47.587 N) / -400.734 N
= 0.012.
Therefore, the coefficient of kinetic friction between the sled and the slope is approximately 0.012.