Question

Consider a 80.0-kg man standing on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.36 m/s in 0.700 s. It travels with this constant speed for the next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.70 s and comes to rest.

I can't seem to get the right answer on this one. Please help. Can someone show me the steps?

Answer 1

Before the elevator starts moving, Reading of the elevator will be only the weight of the man only, i.e.

F= mg= (80)(9.8)= 784 N

Now when elevator starts moving from rest to v=1.36 m/s in time 0.70s, then net acceleration of the elevator will be

a= (v-u)/t= (1.36-0)/0.70= 1.943 m/s²

Let Reading of the scale be "F" then we can write using balance of the force,

F-mg= ma.........................(1)

using all given values in above,

F= m(a+g)= 80.0(1.943+9.8)= 939.44N (ANS)

Now in the constant velocity phase, acceleration of elevator will be 0. Then using equation 1, we'll get reading on the scale as

F-mg= 0

F= mg= (80.0)(9.8)= 784N (ANS)

Now when the acceleration is in negative direction until it comes to rest for time t= 1.70s, then acceleration will be

a'= (0-v)/t'= (0-1.36)/ 1.70= -0.8m/s²

Then using equation 1, reading on the scale will be

  F'-mg= ma'

   F'= m(g+a')= 80.0(9.8+(-0.8))= 720 N (ANS)