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Problem 3: A 60 kg student stands on a bathroom scale in an 740 kg elevator that is supported by a
cable. As the elevator starts moving, the scale reads 480 N.
(a) Find the acceleration of the elevator (magnitude and direction) and the tension in the cable.
[5 Marks]
(b) If the scale reads 660 N, what is the acceleration of the elevator (magnitude and direction)?
[3 Marks]
(c) If the scale reads zero, should the student worry? Explain. [2 Marks]

Sagot :

183so2

Answer:

(a) Finding the acceleration and tension in the cable

To find the acceleration of the elevator and the tension in the cable, we can use Newton's second law of motion.

Given information:

- Mass of the student: 60 kg

- Mass of the elevator: 740 kg

- Scale reading: 480 N

Step 1: Calculate the normal force on the student.

Normal force = Scale reading = 480 N

Step 2: Calculate the net force on the elevator.

Net force = Normal force - Weight of the student

Net force = 480 N - (60 kg × 9.8 m/s²) = 480 N - 588 N = -108 N

Step 3: Calculate the acceleration of the elevator.

Acceleration = Net force / Total mass

Acceleration = -108 N / (740 kg + 60 kg) = -0.133 m/s²

The acceleration of the elevator is -0.133 m/s², which means it is accelerating downward.

Step 4: Calculate the tension in the cable.

Tension in the cable = Total weight - Net force

Tension in the cable = (740 kg + 60 kg) × 9.8 m/s² + 108 N = 7,848 N

(b) Acceleration when the scale reads 660 N

Given information:

- Scale reading: 660 N

Step 1: Calculate the net force on the elevator.

Net force = Scale reading - Weight of the student

Net force = 660 N - (60 kg × 9.8 m/s²) = 660 N - 588 N = 72 N

Step 2: Calculate the acceleration of the elevator.

Acceleration = Net force / Total mass

Acceleration = 72 N / (740 kg + 60 kg) = 0.089 m/s²

The acceleration of the elevator is 0.089 m/s², which means it is accelerating upward.

(c) Interpretation when the scale reads zero

If the scale reads zero, it means that the normal force on the student is zero. This could happen if the elevator is in free fall, meaning it is accelerating downward at the same rate as the student.

In this case, the student should not worry, as the elevator is not experiencing any additional forces beyond gravity. The student would feel weightless, similar to being in a state of microgravity, but there would be no immediate danger.

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