Forces and Motion Challenging Qs

[Saad]

Two boxes, m1 = 1.0kg with a coefficient of kinetic friction of 0.10, and m2 = 2.0kg with a coeffeicient of kinetic friction of 0.20, are placed on an inclined plane inclined at 30 degrees with the horizontal. If a taut string is connected to the blocks, then what acceleration does each block experience?
Note: Free body diagrams should be useful.

 

[HallsofIvy]

Have you made no attempt at all on this? Please look at the "read this before posting" for forum "rules"- basically that you make some attempt yourself and show us what you have tried. That way we have a better idea of what you know and can give suggestions without just giving the answer.

Here are a few hints: you know the mass of each box so you can find its weight and then the component of force down the slope as well as perpendicular to the slope. Use the "coefficient of kinetic friction" with the component of force perpendicular to the slope to get the friction force back up the slope. Now you can find the net force on each box and so could calculate the acceleration of each box down the slope IF they were not tied together by the string.

Since they are tied together by the string, you need to calculate the total force on the TWO together (there is tension in the string but that force pulls up on one box and down on the other and so can be ignored for the motion of both boxes together) and use the TOTAL mass of both boxes to find their acceleration (of course, both boxes must have the same acceleration as long as they are attached by the string).

 

[Saad]

Does this look right?

Box m2:
Weight of m2 box = mg = (2kg)(9.8) = 19.6N
Fg = 19.8N
Fgx = 19.6Nsin30 = 9.8N
Fgy = 19.6Ncos30 = 17N

Box m1:
Weight of m1 box = mg = (1kg)(9.8) = 9.8N
Fg = 9.8N
Fgx = 9.8sin30 = 4.9N
Fgy = 9.8cos30 = 8.5N

Both boxes in a system:
Fgx = 4.9N + 9.8N = 14.7N
Fgy = 17N + 8.5N = 25.5N
Thus: Fgy = Fn (normal force on the system)
****not sure if Fgy is equal to Fn****

Force of Kinetic Friction on box m1:
Fk = (coefficient of Fk)(Fn) = (0.1)(25.5N) = 2.55N

Force of Kinetic Friction on box m2:
Fk = (0.2)(25.5N) = 5.1N

Fnet of m1:
Fnet = Fk + Ft = ma
2.55N + Ft = 1a

Fnet of m2:
Fnet = Fk – Ft = ma
5.1N – Ft = 2a

Solve for a:
2.55N + Ft = 1a
5.1N – Ft = 2a
Since Ft is canceled out:
7.65N = 3a
2.55m/s^2 = a