a 18.0kg sled being pulled across horizontal surface at constant velocity. pulling force has magnitude of 82.0N and is directed at angle of 30 degree above horizontal. determine coefficient of kinetic friction
Uk = (friction force)/(normal force)
at constant velocity
The friction force is the horizontal component of the twowing force.
Uk = (82 cos30)/(Mg - 82 sin 30)
= 71.1/(176.4 - 41.0) = 0.525
To determine the coefficient of kinetic friction, we need to first analyze the forces acting on the sled.
Given:
Mass of sled (m): 18.0 kg
Pulling force (F): 82.0 N
Angle of pulling force (θ): 30 degrees
We know that the sled is moving at a constant velocity, which means the net force acting on it is zero. This implies that the force of friction acting in the opposite direction is equal in magnitude and opposite in direction to the pulling force.
We can break down the pulling force into its horizontal and vertical components. The horizontal component of the force is given by Fcos(θ), and the vertical component is given by Fsin(θ).
Horizontal force:
Fh = F * cos(θ)
Fh = 82.0 N * cos(30 degrees)
Fh = 82.0 N * 0.866
Fh = 71.212 N
Now, we can equate the horizontal component of the pulling force to the force of friction:
Fh = μ * N
Where:
μ is the coefficient of kinetic friction
N is the normal force
Since the sled is on a horizontal surface, the normal force is equal to the weight of the sled, which is given by:
N = m * g
Where:
m is the mass of the sled
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the values:
N = 18.0 kg * 9.8 m/s^2
N = 176.4 N
Now we can solve for the coefficient of kinetic friction:
μ * N = Fh
μ * 176.4 N = 71.212 N
μ = 71.212 N / 176.4 N
μ ≈ 0.403 (rounded to three decimal places)
Therefore, the coefficient of kinetic friction for the sled being pulled across the horizontal surface is approximately 0.403.