A student whose normal weight is 500 N stands on a scale in anelevator and plots

Question

A student whose normal weight is 500 N stands on a scale in anelevator and plots the scale reading on the y axis of agraph. The x axis is time and is divided into 5 secintervals. From 0-5 the scale reads 500N   At 5 sec thescale reading goes to 700N and remains constant at 700N from 5 - 10sec At 10 sec the scale reading drops to 500N and remains constantat 500N from 10- 15 sec   At 15sec the scale readingdrops to 300N and remains steady at 300N ffrom 15 - 20 sec. At 20 sec the scale reading rises to 500N Describe what is happening to student during each 5 secinterval Calculate acceleration of elevator for each 5 secinterval Determine velocity of elevetor at end of each 5 secinterval Determine displacement x of elevator above starting point atend of each 5 second interval A student whose normal weight is 500 N stands on a scale in anelevator and plots the scale reading on the y axis of agraph. The x axis is time and is divided into 5 secintervals. From 0-5 the scale reads 500N   At 5 sec thescale reading goes to 700N and remains constant at 700N from 5 - 10sec At 10 sec the scale reading drops to 500N and remains constantat 500N from 10- 15 sec   At 15sec the scale readingdrops to 300N and remains steady at 300N ffrom 15 - 20 sec. At 20 sec the scale reading rises to 500N Describe what is happening to student during each 5 secinterval Calculate acceleration of elevator for each 5 secinterval Determine velocity of elevetor at end of each 5 secinterval Determine displacement x of elevator above starting point atend of each 5 second interval

Explanation / Answer

Again, please specify which parts you need help with. In general, when the actual weight = apparent weight, the elevatoris not acceratating. This means it is either not moving (0-5 secondinterval, and at 20s) or moving at a constant velocity (10-15sinterval). When the elevator is accelerating up, the apparent weight willincrease (5-10 s interval). And when the elevator is decelerating, the apparent weight willdecrease (15-20 s interval). The formula is that the Apparent Weight = ma - mg (Positive isup) You know his mass (Calcutate from his weight of 500 N), g=~9.8, andyou know his apparent weight for all of the intervals. Should befairly straight forward. For the later parts you will need the two equations of motion, butagain just plug in what you have.

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