1. An elevator (mass 4700 kg ) is to be designed so that the maximum acceleratio

Question

1. An elevator (mass 4700 kg ) is to be designed so that the maximum acceleration is 0.0690 g .

Part A

What is the maximum force the motor should exert on the supporting cable?

Part B

What is the minimum force the motor should exert on the supporting cable?

2. Superman must stop a 100-km/h train in 160 mto keep it from hitting a stalled car on the tracks. The train's mass is 3.6 × 105 kg.

Part A

Determine the force that must be exerted on the train.

Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the force is in the direction of the initial velocity and negative value if the direction of the force is in the direction opposite to the initial velocity.

Part B

Compare the magnitudes of the force exerted on the train and the weight of the train (give as %).

Express your answer using two significant figures.

Explanation / Answer

1.

Part A:

Maximum tension is needed to accelerate the elevator up is
T – 4700*g = 4700(0.0690*g)
T = 4700g + 324.3g

T = 49238.14 N
T = 4.92 x 104 N

Part B:
minimum tension is

T = 4700g - 324.3g

T=42881.86 N

T = 4.28 x 104 N

(2)

Part A

v = 100 km/h = 27.78 m/s

KE = ½*M*v2 = W = F*d

0.5*(3.6 x 105 kg) x (27.78 m/s) 2 = F x 160 m
F = 868055.55 N

F = 8.6805 x 105 N

F = 8.7 x105 N

Part B:

The weight of the train = M x g = 3528000 N

To calculate a percent (868055.55 N / 3528000 N) x 100 = 24.6%

ratio of magnitude = 24.6%

ratio of magnitude = 25% (in two significant figure)

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