Question

An apparatus to determine coefficients of friction is shown. At the angle θ shown with the horizontal, the block of mass m just starts to slide. The box then continues to slide a distance d at which point it hits the spring of force constant k, and compresses the spring a distance x before coming to rest. In terms of the given quantities and fundamental constants, derive an expression for each of the following.

a. µ_s the coefficient of static friction.

b. ∆E, the loss in total mechanical energy of the block-spring system from the start of the block down the incline to the moment at which it comes to rest on the compressed spring.

c. µ_k,  the coefficient of kinetic friction.

Answer 1

(a) We use F_net = 0 for the initial brink of slipping point.

F_gx – f_k = 0 

mg sinθ = μ_s(F_n) 

mg sinθ = μ_smgcosθ 

μ_s = tanθ

(b) Note: we cannot use the friction force from part a since this is the static friction force, we would need kinetic friction. So instead we must apply W_nc = energy loss = ∆K + ∆U + ∆U_sp.∆K is zero since the box starts and ends at rest, but there is a loss of gravitational U and a gain of spring U so those two terms will determine the loss of energy, setting final position as h = 0. Note that the initial height would be the y component of the total distance traveled (d + x) so h = (d+x)sinθ 

U_f – U_i + U_sp(F) – U_sp(i) 

0 – mgh + ½ k∆x² – 0 

½ kx² – mg(d+x)sinθ

(c) To determine the coefficient of kinetic friction, plug the term above back into the work-energy relationship, sub in –Work of friction as the work term and then solve for μ_k  

W_NC = ½kx² – mg(d+x)sinθ 

– f_k(d+x) = ½ kx²  – mg(d + x)sinθ 

- μk mgcosθ(d + x) = ½kx² – mg(d + x)sinθ 

 μk = [mg(d + x)sinθ – ½kx²]/[mg(d + x)cosθ]