An object, previously at rest, slides #5 m# down a ramp, with an incline of #(3pi)/8 #, and then slides horizontally on the floor for another #2 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Question

An object, previously at rest, slides #5  m# down a ramp, with an incline of #(3pi)/8 #, and then slides horizontally on the floor for another #2 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Benjamin Asiedu · Accepted Answer

Start with a force diagram for the object both on the incline and on the straight section. Considering the incline section, and applying Newton's 2nd Law of motion in the direction parallel to the incline, together with the definition of friction, we get : #sumF=ma# #thereforemgsintheta-mu_kmgcostheta=ma# #therefore a=g(sintheat-mu_kcostheta)...............1# Now using energy considerations on the inclined section, we note that potential energy at the top less work done by friction in sliding to the bottom equals left over kinetic energy at the bottom. #therefore mgh-f_kx=1/2mv^2# #therefore mg(5sintheta)-5mu_kmgcostheta=1/2mv^2# #therefore5g(sintheta-mu_kcostheta)=1/2v^2................2# Now using energy considerations along the flat section, application of the Work-Energy Theorem yields : #W_f=DeltaE_k# #thereforemu_kmg*2=1/2mv^2# #therefore 2mu_kg=1/2v^2....................3# Now comparing equations 2 and 3 we observe that they are equal and hence #2mu_kg=5g(sintheta-mu_kcostheta)# #therefore mu_k=(5sintheta)/(2+5costheta)# #=(5sin(3pi/8))/(2+5cos(3pi/8))# #=1.18# Since this value of #mu_k !in [0,1]#, it implies the question is unrealistic and does not happen in practice as proposed in the question.
(Unless I made a mistake somewhere in my calculations - I request other contributors to please check for me).

Nicholas Agrusa · Answer

undefined To find the kinetic friction coefficient, we first need to determine the total work done on the object. The work done against friction on the ramp and on the floor is equal to the change in kinetic energy of the object.

The work done on the ramp can be calculated using the formula: 
$$W_{\text{ramp}} = \text{Force}_{\text{ramp}} \times \text{Distance}_{\text{ramp}} \times \cos(\theta)$$
where:
- $\text{Force}_{\text{ramp}}$ is the force of gravity acting on the object (its weight),
- $\text{Distance}_{\text{ramp}}$ is the distance the object travels on the ramp,
- $\theta$ is the angle of the incline.

The work done on the floor is simply:
$$W_{\text{floor}} = \text{Force}_{\text{floor}} \times \text{Distance}_{\text{floor}}$$
where:
- $\text{Force}_{\text{floor}}$ is the horizontal force of friction,
- $\text{Distance}_{\text{floor}}$ is the distance the object travels on the floor.

Since the object starts from rest and its final velocity is 0, the total work done on the object is equal to the initial kinetic energy of the object, which is 0.

Setting the total work done equal to 0, we can solve for the coefficient of kinetic friction ($μ_k$) using the equation:
$$W_{\text{ramp}} + W_{\text{floor}} = 0$$

Substitute the expressions for $W_{\text{ramp}}$ and $W_{\text{floor}}$ and solve for $μ_k$.