A 9.6 kg box slides down a ramp inclined at 32° to the horizontal. If it accelerates at a rate of 2.9m/s2 , what is the coefficient of kinetic friction, μk? Lay out your work as shown in the lessons.

Question

A 9.6 kg box slides down a ramp inclined at 32° to the horizontal. If it accelerates at a rate of 2.9m/s2 , what is the coefficient of kinetic friction, μk? Lay out your work as shown in the lessons.

Answer 1

To find the coefficient of kinetic friction μ_k, we can use Newton's second law and the force balance along the inclined plane.

The force balance equation along the inclined plane is given by:

F_net ​= m⋅a

Where:

1. F_net is the net force acting on the box along the inclined plane,
2. m is the mass of the box,
3. a is the acceleration of the box.

The net force along the inclined plane can be expressed as the difference between the gravitational force component parallel to the incline and the frictional force:

F_{net =} F_parallel−F_friction

The gravitational force component parallel to the incline can be calculated as:

F_parallel=m⋅g⋅sin(θ)

Where:

1. g is the acceleration due to gravity (approximately 9.8 m/s²),
2. θ is the angle of the incline.

The frictional force can be expressed as:

F_friction = μ_k​⋅F_normal

Where:

1. F_normal is the normal force acting on the box perpendicular to the incline.

F_normal can be calculated as:

F_normal = m⋅g⋅cos(θ)

Now, we can substitute these expressions into the force balance equation and solve for μ_k​:

m⋅a = m⋅g⋅sin(θ)−μ_k​⋅m⋅g⋅cos(θ)

Solving for μ_k:

μ_k ​= (m⋅g⋅sin(θ)−m⋅a )/ (m⋅g⋅cos(θ)​)

Plug in the given values to calculate the coefficient of kinetic friction μ_k.

μ_{k = 0.275}