course Phy 121
Masses of 5 kg and 6 kg are suspended from opposite sides of a light frictionless pulley and are released. What will be the net force on the 2-mass system and what will be the magnitude and direction of its acceleration?
answer/question/discussion: F = ma = 11kg * 9.8m/s^2 = 107.8 Newtons, the direction of the acceleration is in the positive direction of the 6 kg mass
the two gravitational forces tend to accelerate the system in opposite directions
If you give the system a push so that at the instant of release the 5 kg object is descending at 1.8 meters / second, what will be the speed and direction of motion of the 5 kg mass 1 second later?
answer/question/discussion: distance = 1.8m/s *1s = 1.8m
The acceleration of the system is not zero, so the velocity doesn't remain constant during this interval.
Work done by gravity = 1.8m *9.8m/s^2 = 17.64Joules =KE
work is force * displacement; your calculation is displacement * acceleration
The units of m * m/s^2 are m^2 / s^2, not Joules. A Joule is a kg * m ^2 / s^2. You left out the mass.
You also need to consider the fact that one mass moves up while the other moves down.
The work done by the net force is `dKE, the change in KE
KE0 = .5mv0^2 = .5*5kg*(1.8m/s)^2 = 8.1Joules
Kef = `dKE + KE0 = 17.64Joules + 8.1Joules = 25.74Joules
Vf = sqroot(25.74joules/5kg) = 2.27m/s
Good conception of the energy situation. The details aren't right, but the overall approach on this part is.
Direction of motion is positive
During the first second, are the velocity and acceleration of the system in the same direction or in opposite directions, and does the system slow down or speed up?
answer/question/discussion: the acceleration and velocity are in the same direction, the system is speeding up
cq1_12_1 solution and discussion
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Masses of 5 kg and 6 kg are suspended from opposite sides of a light
frictionless pulley and are released.
- What will be the net force on the 2-mass system and what will be the
magnitude and direction of its acceleration?
Let the positive direction be the one in which the 6 kg mass is descending.
Note that the weights of the masses are about 6 kg * 9.8 m/s^2 = 59 N and 5 kg *
9.8 m/s^2 = 49 N.
Two forces thus act on the system, the 59 N force in the positive direction and
the 49 N force in the opposite direction, giving us a net force of about
-
F_net = 59 N - 49 N = 10 N.
The mass of the system is 11 kg. A net force of 10 N on a system of mass 11 kg
results in acceleration
-
a = F_net / m = 10 N / (11 kg) = .9 m/s^2, approximately.
If you give the system a push so that at the instant of release the 5 kg
object is descending at 1.8 meters / second, what will be the speed and
direction of motion of the 5 kg mass 1 second later?
If the 5 kg mass is descending at 1.8 m/s, then the velocity of the system is
-1.8 m/s (note that all directional quantities must be referenced to our
original choice of positive direction).
-
An acceleration of .9 m/s^2 means that in 1 second the velocity of the system
will change by `dv = a `dt = .9 m/s^2 * 1 s = .9 m/s. This will give us a
velocity of vf = v0 + `dv = -1.8 m/s + .9 m/s = -.9 m/s. That is, the 5 kg
object will still be descending, but at .9 m/s rather than at 1.8 m/s.
During the first second, are the velocity and acceleration of the system in
the same direction or in opposite directions, and does the system slow down or
speed up?
During this 1-second interval the acceleration is positive and the velocity
remains negative. So velocity and acceleration are in opposite directions.
Whenever this is the case the object is slowing down, as it clearly is in this
example.
If another second passes then the object's velocity will
be near zero. After reaching the state of rest for an instant, the
continuing acceleration will result in a positive velocity (6 kg mass
descending), which with the positive acceleration will then begin speeding
the system up.