Given the question and answers, answer question E below A mass MI slides on a 45° inclined plane of height H as shown in Fig. 2-14. It is connected by a flexible cord of negligible mass over a small pulley (neglect its mass) to an equal-mass M2 hanging vertically as shown. The length of the cord is such that the masses can be held at rest both at height H/2. The dimensions of the masses and the -pulley are negligible compared to H. At time t=0 the two masses are released. (a) Fort > a calculate the vertical acceleration a of Nh. (b) Which mass will move downward? (c) At what time h will the mass identified in part b) strike the ground? (d) If the mass identified in part b) stops when it hits the ground, but the other mass keeps moving, will it strike the pulley? M₂ 45° M₂ H/2 ANSWERS: A) 1.43 m/s2 downward direction B) mass M2 is going to move down and mass M1 is going to move along the incline in the upward direction C) 0.84H D) the mass M2 strike the pulley it can only move to a vertical height of 0.93H QUESTION TO ANSWER: E) How much force is needed to be applied to M2 for M1 to break the tension in the pulley and fly across to the jungle 3 km away if mass 2 = 25 kg while mass 1 = 20 kg?

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Given the question and answers, answer question E below
A mass MI slides on a 45° inclined plane of height H as shown in Fig. 2-14. It is
connected by a flexible cord of negligible mass over a small pulley (neglect its mass) to
an equal-mass M2 hanging vertically as shown. The length of the cord is such that the
masses can be held at rest both at height H/2. The dimensions of the masses and the
-pulley are negligible compared to H. At time t=0 the two masses are released.
(a) Fort > a calculate the vertical acceleration a of Nh.
(b) Which mass will move downward?
(c) At what time h will the mass identified in part b) strike
the ground?
(d) If the mass identified in part b) stops when it hits the
ground, but the other mass keeps moving, will it strike the
pulley?
ANSWERS:
M₂
45°
M₂
D)
the mass M2 strike the pulley it can only move to a vertical height of 0.93H
QUESTION TO ANSWER:
H/2
A) 1.43 m/s2 downward direction
B) mass M2 is going to move down and mass M1 is going to move along the incline in the upward
direction
C) 0.84H
E) How much force is needed to be applied to M2 for M1 to break the tension in the pulley and fly
across to the jungle 3 km away if mass 2 = 25 kg while mass 1 = 20 kg?
Transcribed Image Text:Given the question and answers, answer question E below A mass MI slides on a 45° inclined plane of height H as shown in Fig. 2-14. It is connected by a flexible cord of negligible mass over a small pulley (neglect its mass) to an equal-mass M2 hanging vertically as shown. The length of the cord is such that the masses can be held at rest both at height H/2. The dimensions of the masses and the -pulley are negligible compared to H. At time t=0 the two masses are released. (a) Fort > a calculate the vertical acceleration a of Nh. (b) Which mass will move downward? (c) At what time h will the mass identified in part b) strike the ground? (d) If the mass identified in part b) stops when it hits the ground, but the other mass keeps moving, will it strike the pulley? ANSWERS: M₂ 45° M₂ D) the mass M2 strike the pulley it can only move to a vertical height of 0.93H QUESTION TO ANSWER: H/2 A) 1.43 m/s2 downward direction B) mass M2 is going to move down and mass M1 is going to move along the incline in the upward direction C) 0.84H E) How much force is needed to be applied to M2 for M1 to break the tension in the pulley and fly across to the jungle 3 km away if mass 2 = 25 kg while mass 1 = 20 kg?
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