A 4.0 kg toboggan rests on a frictionless icy surface, and a 2.0 kg block rests on
top of the toboggan. The coefficient of static friction between the block and
the surface of the toboggan is 0.60, whereas the kinetic friction coefficient is
0.51. The block is pulled by a 30 N-horizontal force as shown. What are the
magnitudes and directions of the resulting accelerations of the block and the
Fnet = F-Ffr = ma
Ffr = μk*Fn
The Attempt at a Solution
I found that the box will move ontop of the toboggan because μs *Fn = 11.76 N which is less than 30 N so the box will move.
F-Ffr = ma
30N - 10.0N = 2.0 kg (a)
accel. = 10 m/s^2
so i know thats the box on the toboggan. How do i find the toboggan on the ground?
do i just do it as
F-Fr = (Mb +Mt) *a
30 - 10 = (2.0 + 4.0) *a
20 = 6.0 *a
a = 3.33m/s^2
does that make sense at all? any help would be great.