Sled Sliding Down a Hill....Conservation Energy, Help?

Question

Sled slides down a hill w/ snow with no initial velocity. The hill is 20 meters high. The total distance traveled on the x axis is 22 meters. (So the length of the hill's slope is 29.7 meters, if thats relevant.) The coefficient of kinetic friction between the sled and snow is .353 . The student and sled together weight 71 kg. What is the final velocity at the bottom of the hill?

Part 2
After the hill there is a flat area 4 meters across. Does the sled travel this entire distance? If so how fast are they going at the end. If not , how far do the only make it?

Part 3(if it makes it past flat area)
Then after the flat area, the sled comes up to another hill. The problem does not give me an angle of the hill, just that it is 7 meters across the x axis, and i'm supposed to find how far the sled goes up that hill?

Answer 1

Part 1

The potential energy is reduced
=m*g*h

Some energy gets converted to heat due to friction

The frictional energy is
=m*g*cos(th)*u*L
where th is the angle of the hill and u is the coefficient of friction and L is the length of the slope

there was no initial kinetic energy, so the final kinetic energy
.5*m*v^2
is the loss in potential energy less the frictional work'
.5*m*v^2=m*g*h-m*g*cos(th)*u*L

the mass divides out
v^2=2*g*(h-cos(th)*u*L)
=2*9.81*(20-22*.353)
note that cos =x/hypotenutse=22/L

v^2=240
v=15.5 m/s

Part 2

the energy lost on the 4 m flat is
m*g*u*4
so the kinetic energy at the end of the run
.5*m*v^2=.5*m*15.5^2-m*g*u*4
v^2=15.5^2-2*9.81*.353*4

v=14.58 m/s

Part 3
For this you also need the angle of the hill

I will give you the solution using variables

When the sled comes to a stop up the hill the intial kinetic energy will be converted to potential energy m*g*h, and frictional heat loss m*g*u*cos(th)*L

cos(th) = x/L
so friction is m*g*u*x

.5*m*14.58^2=m*g*(h-u*x)
h-u*x=14.58^2/(2*g)

If you knew the angle you could relate the tan(th) to x and h

tan(th)=h/x

and solve for the two variables.

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