# A slide-loving pig slides down a certain 35 degree slide in twice the time it would take to slide down a frictionless 35 degree slide. What is the coefficient of kinetice friction between the pig and the slide?

I know that when the coefficent of kinetic fricition is zero t=1/2 and When it is unknown t =1 but I am not sure....

Thinking. If if slides down in twice the time, its average velocity is twice, and its final velocity is twice as large. If its final veloicity is twice, the Kinetic energy is four times as much on the frictionless plane. So if the available energy is mgh, then friction is absorbing 3/4 of that energy, leaving 1/4 for KE at the bottom.

3/4 mgh= mu*m*g*cos35*distance

but distance= h/sin35

and one can solve for mu.

check my thinking. Neat question.

## To solve this problem, you can start by assuming that the pig slides down the frictionless 35-degree slide in time t. This means that the time it takes for the pig to slide down the slide with kinetic friction is 2t (since it's twice the time).

Next, let's consider the work-energy principle. The total work done on an object is equal to the change in its kinetic energy. We can assume that the pig starts from rest, so the initial kinetic energy is zero.

On the frictionless slide, the work done is equal to the change in gravitational potential energy. The gravitational potential energy is given by mgh, where m is the mass of the pig, g is the acceleration due to gravity, and h is the height of the slide.

On the slide with kinetic friction, the work done is equal to the change in gravitational potential energy minus the work done by the friction force. The work done by the friction force is given by the product of the friction force and the distance traveled.

Now, let's calculate the work done on the frictionless slide:

Work_fricless = mgh

Next, let's calculate the work done on the slide with kinetic friction:

Work_fric = mgh - friction * distance

Since the pig slides down the slide in twice the time on the frictionless slide, its average velocity is twice as large. This also means that its final velocity is twice as large, resulting in four times the kinetic energy at the bottom of the slide on the frictionless slide.

We can equate the work done on the frictionless slide to four times the work done on the slide with kinetic friction:

4 * Work_fricless = Work_fric

Substituting the expressions for work, we get:

4 * mgh = mgh - friction * distance

Now, we need to express the distance traveled in terms of h. We can use trigonometry to find that the distance is equal to h/sin(35 degrees).

Substituting this into the equation and solving for friction:

4 * mgh = mgh - friction * (h / sin(35 degrees))

Dividing both sides by mgh and simplifying, we get:

3 = friction / sin(35 degrees)

Finally, multiplying both sides by sin(35 degrees), we get:

friction = 3 * sin(35 degrees)

So the coefficient of kinetic friction between the pig and the slide is approximately equal to 0.988.

Keep in mind that this is an approximation, and you may need to consider additional factors such as air resistance or other sources of friction if they are relevant to the problem.