A 14.4-kg sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 66.9 N and is directed at an angle of 32 ° above the horizontal. Determine the coefficient of kinetic friction
M*g = 14.4 * 9.8 = 141.1 N.=Wt. of sled.
Fn = 141.1 - 66.9*sin32 = 105.7 N. =
Normal force.
Fap*Cos A-Fk = M*a.
66.9*Cos32-Fk = M*0 = 0.
Fk = 56.7 N. = Force of kinetic friction.
u = Fk/Fn.
To determine the coefficient of kinetic friction, we need to first find the force of kinetic friction acting on the sled.
The force of kinetic friction can be calculated using the equation:
F_friction = μ * N
Where:
F_friction is the force of friction (unknown)
μ is the coefficient of friction (unknown)
N is the normal force (depends on the weight of the sled)
To find the normal force N, we use the equation:
N = mg
Where:
m is the mass of the sled (14.4 kg)
g is the acceleration due to gravity (9.8 m/s^2)
Substituting the values, we have:
N = (14.4 kg) * (9.8 m/s^2)
N = 141.12 N
Next, we need to resolve the pulling force into horizontal and vertical components.
F_pull_horizontal = F_pull * cos(θ)
F_pull_vertical = F_pull * sin(θ)
Where:
F_pull is the pulling force (66.9 N)
θ is the angle of the pulling force (32°)
Substituting the values, we have:
F_pull_horizontal = (66.9 N) * cos(32°)
F_pull_horizontal = 56.50 N
Now we can calculate the force of kinetic friction acting on the sled:
F_friction = μ * N
Since the sled is being pulled at a constant velocity, the force of kinetic friction is equal in magnitude and opposite in direction to the horizontal component of the pulling force:
F_friction = F_pull_horizontal
Therefore:
F_friction = 56.50 N
Finally, we can determine the coefficient of kinetic friction:
F_friction = μ * N
Substituting the values, we have:
56.50 N = μ * 141.12 N
Solving for μ:
μ = 56.50 N / 141.12 N
μ ≈ 0.40
Therefore, the coefficient of kinetic friction is approximately 0.40.
To determine the coefficient of kinetic friction, we first need to understand the forces acting on the sled.
In this scenario, the sled is being pulled at a constant velocity. This means that the forces acting on the sled are in equilibrium (balanced forces), resulting in zero net force. From this, we can conclude that the force of friction (opposing the motion) is equal in magnitude to the applied pulling force.
The pulling force can be resolved into its horizontal and vertical components using trigonometry. The vertical component does not affect the horizontal motion of the sled because it is perpendicular to the horizontal surface. Hence, we only need to consider the horizontal component of the pulling force.
The horizontal component of the pulling force can be calculated as follows:
F_horizontal = F_applied x cos(θ)
where F_applied is the given pulling force with a magnitude of 66.9 N and θ is the given angle of 32°.
F_horizontal = 66.9 N x cos(32°)
F_horizontal ≈ 56.313 N
Now, let's consider the force of friction. The force of friction can be determined using the equation:
F_friction = coefficient of friction x normal force
Since the sled is on a horizontal surface, the normal force equals the weight of the sled which can be calculated using the formula:
Normal force = mass x gravity
where the mass of the sled is given as 14.4 kg and the acceleration due to gravity is approximately 9.8 m/s².
Normal force = 14.4 kg x 9.8 m/s²
Normal force ≈ 141.12 N
Plugging this back into the equation for the force of friction:
F_friction = coefficient of friction x 141.12 N
Since the applied pulling force and the force of friction are equal in magnitude (56.313 N), we can set up the equation:
56.313 N = coefficient of friction x 141.12 N
Simplifying this equation:
coefficient of friction = 56.313 N / 141.12 N
coefficient of friction ≈ 0.398
Therefore, the coefficient of kinetic friction is approximately 0.398.