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Homework Statement
A 15kg toboggan is pulled across the snow at a constant speed by a horizontal force of 22N.
a) What is the force of gravity on the toboggan?
b) What is ##μ_k##?
c) How much more force is needed to pull the toboggan if 105kg is added to the toboggan?
Homework Equations
##m = 15kg##
##a = 9.8 m/s^2##
##F_A = 22N## [Applied force]
##F_N## [Force normal]
##F_K## [Kinetic force/Friction]
The Attempt at a Solution
a) ##F_G = ma = (15 kg)(9.8 m/s^2) = 147N##
b) The toboggan travels at a constant velocity ( a = 0 ) with no ups and downs. This implies that ##F_N = F_G = 147N## and ##F_K = F_A = 22N##
Now ##F_K = μ_KF_N \Rightarrow μ_K = \frac{F_K}{F_N}##
So that ##μ_K = \frac{22N}{147N} = 0.15##.
c) Hm so the total weight of the toboggan will now be ##m = 120kg## and we have the kinetic friction coefficient ##μ_K = 0.15##.
Would it just be :
##F_K = μ_Kma = (0.15)(120 kg)(9.8 m/s^2) = 176.4N##
So we would require 176.4N+22N = 198.4N more to keep the toboggan at a constant velocity. Is this okay?