How can I find kinetic friction in this problem? What equation would work?

In the figure below, a slide loving pig slides down a certain 27掳 slide in twice the time it would take to slide down a frictionless 27掳 slide. What is the coefficient of kinetic friction between the pig and the slide? How can I find kinetic friction in this problem? What equation would work?
a=acceleration with friction


A=acceleration without friction


s=1/2At^2


s=1/2a(2t)^2


.5At^2=2at^2


a=1/4A


Fnet=mA


mgsin(pheta)=mA


A=gsin(pheta)


Fnet=ma


Fg-f=ma


mgsin(pheta)-umgcos(pheta)=ma


gsin(pheta)-ugcos(pheta)=1/4A


gsin(pheta)-ugcos(pheta)=1/4(gsin(phet鈥?br>

ugcos(pheta)=3/4gsin(pheta)


u=3/4tan(pheta)


u=3/4tan(27)


u=.382





make it a good day


How can I find kinetic friction in this problem? What equation would work?
The equation to use is actually F=ma and d = 0.5*a*t^2. There is only one force, gravity. The component parallel to the slop is the force that is accelerating the pig. The difference between the two cases is friction, so F2 = C * F1, where F2 = force with friction, F1 = force without friction, and C = coefficient.





We know the distance travelled in both case are the same, the time is different. Therefore:





a1 = 2*d/t1^2 and a2 = 2*d/t2^2





m * a2 = C * m * a1


C = a2/a1 = (2*d2/t2^2) / (2*d1/t1^2) = t1^2/t2^2 = (t1/t2)^2





We are given that with friction it took twice as long, therefore:





t1/t2 = 0.5


C = (t1/t2)^2 = (0.5)^2 = 0.25