✔ 最佳答案
Since m2 is moving up the inclined plane, the net force on m2 is equal to the tension in the string minus the sum of the component of its weight that is parallel to the inclined plane and the friction force
Force parallel = m2 * g * sin θ
sin θ = 5/13
Force parallel = 15 * 9.8 * 5/13 = 735/13
This is approximately 56.54 N.
Ff = μk * m2 * g * cos θ
cos θ = 12/13
Ff = 0.19 * 15 * 9.8 * 12/13 = 335.16/13
This is approximately 25.78 N.
Total force = 735/13 + 335.16/13 = 82.32 N
Net force = T – 82.32 = 15 * a
T = 82.3 + 15 * a
The free body diagram of m1 show 2 parts of rope above it. When m1 is not moving, each side the string is supporting one half of m1’s weight. Since one end of string is attached to the ceiling, this end will not move.
Weight = 38 * 9.8 = 372.4 N
Net force = 372.4 – 2 * T = 38 * a
2 * T = 372.4 – 38 * a
T = 186.2 – 19 * a
Let’s set these two equations equal to each other and solve for a.
82.3 + 15 * a = 186.2 – 19 * a
34 * a = 186.2 – 82.3
a = 103.9 ÷ 34
This is approximately 3.056 m/s^2.
186.2 – 82.3 is one half of the weight of m1 minus the sum of the other two forces. This would have been a much easier way to determine the acceleration.
2) If the system is released from rest, what is the speed of mass m2 after 3.5 s ?
Use the following equation.
vf = vi + a * t, vi = 0
vf = (103.9 ÷ 34) * 3.5
This is approximately 10.7 m/s.
3) When mass m2 moves a distance 4.12 m up the ramp, how far downward does mass m1 move?
As m2 moves a distance 4.12 m up the ramp, 4.12 meters of the string will move over the pulley. Then down to the second pulley. This will cause m1 to move 2.06 meters downward. The other 2.06 meters is on the opposite of the second pulley.
