A block is on an inclined plane with a string attached from the block to a pulley at the top of the incline.
The coefficient of kinetic friction between the mass and the plane is 0.1. The mass is 100 kg. The plane is inclined at 30 degrees.
If the mass is moving up the inclined plane at a velocity of 2 m/s, what tension should be applied to the rope in order to make the mass reverse directions in exactly 1 s?
My method of solving this:
I first set up my equation as:
(force tension) - (force gravity x) - (force friction) = ma
Since acceleration will abruptly change in the downward direction, solving for a:
v final = v + at
0 = 1 + a (1 second)
a = -1 m/s^2 ---> since its going down the slope, sign will change to +
Plugging back into first question and solving for force of tension
force tension = [100 kg (10 m/s^2) sin 30] + [0.1 (100)(10) cos 30] + 100 (1 m/s^2)
Ft = 685 N
However, this is incorrect. The answer is 387 N.
I'm stumped how to solve this problem. I'd appreciate any help on this, thanks!
The coefficient of kinetic friction between the mass and the plane is 0.1. The mass is 100 kg. The plane is inclined at 30 degrees.
If the mass is moving up the inclined plane at a velocity of 2 m/s, what tension should be applied to the rope in order to make the mass reverse directions in exactly 1 s?
My method of solving this:
I first set up my equation as:
(force tension) - (force gravity x) - (force friction) = ma
Since acceleration will abruptly change in the downward direction, solving for a:
v final = v + at
0 = 1 + a (1 second)
a = -1 m/s^2 ---> since its going down the slope, sign will change to +
Plugging back into first question and solving for force of tension
force tension = [100 kg (10 m/s^2) sin 30] + [0.1 (100)(10) cos 30] + 100 (1 m/s^2)
Ft = 685 N
However, this is incorrect. The answer is 387 N.
I'm stumped how to solve this problem. I'd appreciate any help on this, thanks!