Question

V. ELEVATOR REVIEW A man stands on a scale in an elevator. The elevator moves downwards at a velocity of 10 m/s. The elevator accelerates “pwards at a rate of 5 m/s.. The scale under the man reads 180 lbs. A. Is the elevator slowing down or speeding up? B. Given that 1 pound 4.5 Newtons, compute the man's mass? HINT: normal force is not necessarily cqual to weight. You have to usc NII C. The elevator reaches a complete stop and begins moving upwards, stl moving with the same (constant) acceleration of 5 m/s. What will happen to the reading on the scale? Will it go upwards, downwards, or stay the same? D. Suddenly, the elevator begins to accelerate downwards. What will happen to the reading on the scale? Will it go upwards, downwards, or stay the same? E. What downward acceleration must the elevator achieve in order for the scale to read 0?

Answer 1

Initial velocity of elevator $u=-10\,m/s$ (Downward is taken negative.)

Acceleration of elevator $a=5\,m/s^2$ ( upward is taken positive)

A) Final velocity of elevator $v=u+at$

$v=-10+5t$

at time $t=0\,s$ , $v=u=-10\,m/s$

at time $t=2\,s$ , $v=-10+10=0$

That is the elevator is slowing down.

B)

Net force on elevator in the vertical direction , $N=m(g+a)$

Reading on scale $N=180\,lbs=810\,N$

$m(g+a)=810$

$m(9.81+5)=810$

$m=54.7\,kg=120.6\,lb$

C ) Reading o the scale $N=m(g+a)=54.7(9.81+5)=810\,N$ , so the reading on the scale stay the same.

D ) Elevator begins to accelerate downwards, $N=m(g-a)$

Reading on the scale decreases.

E ) To reading on the scale is zero.

$N=m(g-a)=0$

$m(g-a)=0$

Downward acceleration is $a=g=9.81\,m/s^2$