Question

A 20 kg sled is being pulled across a horizontal force at a constant velocity the pulling force has a magnitude of 80 N and it's directed at an angle of 30° above the horizontal determine the coefficient of kinetic friction

Answer 1

The coefficient of kinetic friction is found by resolving the applied force into horizontal and vertical components, determining the reduced normal force due to the vertical component, and equating the kinetic friction force to the horizontal component of the applied force.

Step-by-step explanation:

To determine the coefficient of kinetic friction when a 20 kg sled is pulled across a horizontal surface at constant velocity with an applied force of 80 N at a 30° angle, we can use the fact that the net force in the horizontal direction is zero.

First, we resolve the 80 N force into horizontal and vertical components. The horizontal component of the force (Fhorizontal) is 80 N × cos(30°) and the vertical component (Fvertical) is 80 N × sin(30°). The vertical component acts against gravity and reduces the normal force exerted by the surface on the sled. Hence, the normal force (N) is reduced to the sled's weight minus Fvertical, expressed as mg - 80 N × sin(30°), where m is the mass of the sled and g is the acceleration due to gravity (9.8 m/s²).

Since the sled is moving at a constant velocity, the kinetic friction force (f-k) must be equal and opposite to the horizontal component of the applied force. Therefore, f-k = Fhorizontal. The kinetic friction force is also defined as f-k = μk N, where μk is the coefficient of kinetic friction. By equating the two expressions for f-k and solving for μk, we obtain the coefficient of kinetic friction.

Answer 2

0.58

Step-by-step explanation:

Using the frictional force formula expressed as;

`F_f = \mu R\\`

`\mu` is the coefficient of kinetic friction

R is the normal reaction

`\mu = (F_f)/(R)`

`F_f = Wsin \theta\\R = W cos\theta\\\mu = (Wsin \theta)/(W cos \theta)\\\mu = (sin\theta)/(cos \theta) \\\mu = tan \theta`

Given

`\theta = 30^0\\\mu = tan 30^0\\\mu = 0.58`

hence the coefficient of kinetic friction is 0.58