1. A kid was tobogganing but the warm sun melted the snow and he must now pull his 10kg toboggan on the muddy slope (coefficient of kinetic friction is 0.3, the bottom of the hill is 10 degrees). He pulls with a force of 30N. The parallel force applied is 20cos30.

At what rate does the toboggan accelerate.

So I first needed to find the normal force for friction.
EF = 0
Fgperp - Fn = 0
100cos10 - Fn = 0 (i got a 100..that's the force of gravity. on the free body diagram i needed to calculate the perpendicular force of gravity)
Fn = -98.48 (down is positive, therefore Fn is negative).

!@#$%^&ara - Ffr = ma
20cos30 - (0.3)(-98.48) = 10a
46.86/10 = a
a = 4.7 m/s^2

b) The toboggan ran onto an ice-covered patch of ground. The kid found that it moved at a constant speed without being pulled. Determine the coefficient of kinetic friction between the toboggan and the ice-covered ground.

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asked by Alex
  1. Your acceleration looks too high: almost 1/2 g. Are you using 10 m/s^2 for g? Why is the parallel force applied 20 cos 30? Is 30 the angle between the rope and the hill slope? Why are you multiplying by 20 and not 30 N? Are you taking into account that pulling up on the rope reduces the normal force on the ground?

    I don't get the picture of what is going on and don't understand what you are doing. Your computed acceleration is too high. The problem seems to be your double minus (and therefore +) sign in the friction term. Friction cannot end up adding to the pulling force.

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    posted by drwls

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