kinetic friction?

Question

a slide loving pig slides down a certain 16° slide in twice the time it would take to slide down a frictionless 16° slide. What is the coefficient of kinetic friction between the pig and the slide?

Answer 1

consider the frictionless case using the motion equation
L=.5a*t^2, where L is the length of the slide
looking at a fbd of the pig, and F=m*a
F=sin(16)*m*g
so a=sin(16)*g
and L=.5*sin(16)*g*t^2

in the friction case
L is the same, but F is now
sin(16)*m*g-cos(16)*m*g*u
a=g*(sin(16-cos(16)*u)
where u is the friction coefficient

since we know that this time is 2*t the first time
L=.5*g*(sin(16-cos(16)*u)
*4*t^2

since the L is constant
sin(16)=(sin(16-cos(16)*u)*4
u=3*sin(16)/(4*cos(16))
u=.215

j

Answer 2

Draw a free body diagram.
You'll find that mgsinΘ points down the ramp, and umgcosΘ (friction) points up the ramp. u being mu, coefficient of friction
Write an equation for net force:
F = ma = mgsinΘ - umgcosΘ
a = gsinΘ - ugcosΘ if there is friction, and
a = gsinΘ if there is no friction
Since d = at²/2, and d is constant, a needs to be 1/4 times as much to for the pig to slide down in twice the time.
¼gsinΘ = gsinΘ - ugcosΘ
Solve for u, and you get
u = 3tanΘ/4
plug & chug
u = 3tan(16)/4 = 0.215

Answer 3

distance = 1/2 at^2

So time is inversely proportional to square root of a. So the piggy needs four times the acceleration to slide in half the time.

with no friction, a = g sin theta

with friction a = g sin theta - mu g cos theta.

So 1/4 (g sin theta) = g sin theta - mu g cos theta

Solve for mu.

mu = 3/4 tan (theta)