An object, previously at rest, slides 2 m2m down a ramp, with an incline of (pi)/12 π12, and then slides horizontally on the floor for another 24 m24m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

1 Answer
Nov 21, 2017

The coefficient of kinetic friction is =0.02=0.02

Explanation:

At the top of the ramp, the object possesses potential enegy.

At the botton of the ramp, part of the potential energy is converted into kinetic energy and the work done by the frictional force.

The slope is s=2ms=2m

The angle is theta=1/12piθ=112π

Therefore,

PE=mgh=mg*ssinthetaPE=mgh=mgssinθ

mu_k=F_r/N=F_r/(mgcostheta)μk=FrN=Frmgcosθ

F_r=mu_k*mgcosthetaFr=μkmgcosθ

Let the speed at the bottom of the ramp be (u)ms^-1(u)ms1

so,

mg*ssintheta=1/2m u^2+s*mu_k*mgcosthetamgssinθ=12mu2+sμkmgcosθ

gssintheta=1/2u^2+smu_kgcosthetagssinθ=12u2+sμkgcosθ

u^2=2gs(sintheta-mu_kcostheta)u2=2gs(sinθμkcosθ)

On the horizontal part,

The initial velocity is =u=u

The final velocity is v=0v=0

The distance is d=24md=24m

The deceleration is calculated with Newton's Second Law

F=F'_r=mu_kmg=ma

a=-mu_kg

We apply the equation of motion

v^2=u^2+2ad

0=2gs(sintheta-mu_kcostheta)-2mu_kgd

2mu_kgd=2gssintheta-mu_k2gscostheta

2mu_kgd+mu_k2gscostheta=2gssintheta

mu_k(2gd+2gscostheta)=2gssintheta

mu_k=(gssintheta)/(gd+gscostheta)

=(9.8*2*sin(1/12pi))/(9.8*24+9.8*2*cos(1/12pi))

=5.07/(254.1)

=0.02