(a) We can first find the acceleration of the whole system by considering the net force acting on it due to gravity. The net force acting on the system can be determined by the force acting on the hanging block, m3, which is m3 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
The equations of motion for the blocks are given by Newton's second law:
m1 * a = T
m2 * a = T
m3 * g = T + m1 * a + m2 * a
We can sum the first two equations and substitute in the third equation:
m1 * a + m2 * a = m3 * g - m1 * a - m2 * a
Solve for the acceleration a:
a = m3 * g / (m1 + m2 + m3)
Now we have the acceleration, so we can use the second equation of motion to find the tension T:
T = m2 * a
(a) Tension T:
m1 = 1.75 kg
m2 = 2.85 kg
m3 = 4.95 kg
g = 9.8 m/s^2
a = (4.95 * 9.8) / (1.75 + 2.85 + 4.95)
a ≈ 3.29 m/s^2
T = 2.85 * 3.29
T ≈ 9.38 N
The tension T in the string connecting the two blocks on the horizontal surface is approximately 9.38 N.
(b) The time t required for the hanging weight to fall 10.0 cm can be obtained from the kinematic equation of motion:
y = 1/2 * a * t^2
Since the block is falling, its distance traveled, y, is negative. Convert 10.0 cm to meters:
y = -0.1 m
Solve for t:
t = sqrt(-2 * y / a)
t = sqrt(-2 * (-0.1) / 3.29)
t ≈ 0.39 s
The time required for the hanging weight to fall 10.0 cm, starting from rest, is approximately 0.39 seconds.
Two blocks connected by a string are on a horizontal frictionless surface. The blocks are connected to a hanging weight by means of a string that passes over a pulley as shown in the figure below, where m1 = 1.75 kg, m2 = 2.85 kg, and m3 = 4.95 kg.
(a) Find the tension T in the string connecting the two blocks on the horizontal surface.
(b) How much time is required for the hanging weight to fall 10.0 cm if it starts from rest?
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