Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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.
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.