1. The problem statement, all variables and given/known data 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 μs between the block and the surface of the toboggan is 0.58, whereas the kinetic friction coefficient is 0.48. The block is pulled by a horizontal force of 30 N as shown. a.Calculate the block’s acceleration. b.Calculate the toboggan’s acceleration c.If the applied force is gradually reduced, at what value do the block and the toboggan have the same acceleration? 2. Relevant equations F=ma Ff = ukFn Ff = usFn mb = block's mass mt = toboggan's mass Fa = applied force (30N) 3. The attempt at a solution a. For the block: Fg = mb = (2)(9.8) = 19.6N No acceleration in y direction therefore FN = 19.6N Ff = usFn = (0.58)(19.6N) = 11.37N -- from this I know that the applied force overcomes the max static friction force and the block will indeed move and we need to use kinetic friction. (** Am I correct in including this?) Ff = ukFn = (0.48)(19.6N) = 9.41N F=ma = Fa - Ff= ma 30N- 9.41N = 2a a= 10.30 m/s^2, to the left b. F= mta = Ff = mta 9.41N = 4a a=2.35 m/s^2, also to the left (** Am i correct to use the frictional force of the block here?) c. F(toboggan) = mta = 9.41N/4kg =2.35 m/s^2 Since it is the applied force that is decreasing, I have to get the acceleration of the block to 2.35m/s^2. F(block)=mba Fa-Ff=mba F-9.41=(2kg)(2.35m/s^2) F= 14.11N Just looking to get some advice as to if I'm going through this correctly! It confused me when giving me both the static and kinetic coefficient of frictions, and looking online it seems like everyone was doing this question differently. Thanks!!