A slide loving pig slides down a certain 27° slide in four times the time it wou

Question

A slide loving pig slides down a certain 27° slide in four times the time it would take to slide down a frictionless 27° slide. What is the coefficient of kinetic friction between the pig and the slide? What component of the gravitational force is perpendicular to the slide? How is the normal force related to that component? How is the kinetic frictional force related to the normal force? Is the frictional force up or down the slide? What component of the gravitational force is down the slide? What is the net force along the slide? What is the acceleration? The constant-acceleration equations can be used. How much time is needed?

Explanation / Answer

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(27)*m*g
so a=sin(27)*g
and L=.5*sin(27)*g*t^2

in the friction case
L is the same, but F is now
sin(27)*m*g-cos(27)*m*g*u
a=g*(sin(27-cos(27)*u)
where u is the friction coefficient

since we know that this time is 4*t the first time
L=.5*g*(sin(27-cos(27)*u)
*16*t^2

since the L is constant
sin(27)=(sin(27-cos(27)*u)*16
u=15*sin(27)/(16*cos(27))
u= 0.478

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