During the winter break you go sledding with your friends. The hill you are sled
ID: 1430221 • Letter: D
Question
During the winter break you go sledding with your friends. The hill you are sledding on is L = 11 m long (that's the distance along the slope, not the vertical height of the slope) and slopes at an angle of ? = 29 degrees. At the bottom of the hill is a long horizontal stretch of snow where you slow down and stop.
If you start with zero velocity at the top of the hill and you stop a distance d = 26 m from the bottom of the hill, what is the coefficient of friction, ?k of the sled on the snow? Assume it is the same on both the hill and along the horizontal.
Explanation / Answer
solution:
Friction is times the normal force. ("normal" is an old-fashioned word meaning "perpendicular.")
When you are on the hill, you can project your gravitational force into the rotated coordinate system with an x-axis parallel to the hill, and a y-axis perpendicular to it. The result is a vector:
(g sin 29°, g cos 29°).
Thus, along the hill, your driving force is mg sin 30° and your normal force is mg cos 29°. Your frictional force is just m g cos 29°, and it points opposite to the driving force of gravity.
Work is force times distance. Both of these forces perform work on you through the L1 = 11m distance of the hill, and this work (by the work-energy theorem) becomes your kinetic energy at the base of the hill, before you travel down the horizontal pathway. This work is just:
W1 = (m g sin 30° - m g cos 30°) * L1.
On the bottom of the hill, your normal force increases to be just mg -- the ground beneath you directly opposes gravity, rather than turning it into forward motion like the hill did. And there is no more driving force. If the coefficient of friction is still the same, then the work done by friction over this distance L2 = 26 m is just:
W2 = - m g * L2.
This must remove the kinetic energy that you just acquired, so that we can say:
(m g sin 29° - m g cos 29°) * L1 - m g * L2 = 0.
We can now divide through by m*g to simplify this:
(sin 29° - cos 29°) L1 - L2 = 0.
L1 sin 29° = ( L1 cos 29° + L2 )
= L1 sin 29° / ( L1 cos 29° + L2 )
When I plug in the numbers I get about 0.1497
upto 3signifiacant figure 0.149 ans
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