Block on toboggan pulled by 30 N force on frictionless ice

[NKKM]

HI,
I am confused about how to approach this question.

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 m
s 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 toboggan?

If I calculate the acceleration of the box as so: Fpull- Ffriction = Fnet = ma
using the kinetic friction coefficient and solve for acceleration. Does that make sense.. and how then do I approach the acceleration of the toboggan?

 

[member 553083]

To calculate the acceleration of the block and the toboggan, you can use Newton's Second Law. The net force acting on the block is the sum of the pull force and the friction force. This force will cause an acceleration of the block, which can be calculated using the equation: Fnet = ma, where m is the mass of the block and a is the acceleration. The friction force acting on the block can be calculated by multiplying the coefficient of kinetic friction (0.51) with the normal force. The normal force is equal to the weight of the block, which can be calculated using the equation: Fg = mg, where m is the mass of the block and g is the acceleration due to gravity (9.81 m/s2). Thus, you can calculate the acceleration of the block by substituting the calculated friction force and the pull force into the equation: Fpull + Ffriction = ma. The acceleration of the toboggan can also be calculated using Newton's Second Law. The net force acting on the toboggan is the sum of the pull force and the friction force. The friction force acting on the toboggan can be calculated by multiplying the coefficient of static friction (0.60) with the normal force. The normal force is equal to the sum of the weight of the toboggan and the weight of the block, which can be calculated using the equation: Fg = mg, where m is the mass of the combined system and g is the acceleration due to gravity (9.81 m/s2). Thus, you can calculate the acceleration of the toboggan by substituting the calculated friction force and the pull force into the equation: Fpull + Ffriction = ma. I hope this helps!