A 77.0 kg man stands on a spring scale in an elevator. Starting from rest, the e

Question

A 77.0 kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.25 m/s in 1.00 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 1.80 s and comes to rest.

(a) What does the spring scale register before the elevator starts to move?
? N

(b) What does it register during the first 1.00 s?
? N

(c) What does it register while the elevator is traveling at constant speed?
? N

(d) What does it register during the time it is slowing down?
? N

Explanation / Answer

Fn = force by the spring scale

mg = weight of the man

a)

before the elevator starts to move

force equation for the man is given as

Fn - mg = 0

Fn = mg

Fn = 77 x 9.8

Fn = 754.6 N

b)

Vf = final velocity = 1.25 m/s

Vi = initial velocity = 0 m/s

During 1 sec : acceleration "a" is given as

a = Vf - Vi / t = 1.25 - 0 / 1 = 1.25 m/s2

force equation for the man is given as

Fn - mg = ma

Fn = m (g + a) = 77 (9.8 + 1.25)

Fn = 850.85 N

c)

when elevator moves at constant speed , a =0

so force equation for the man is given as

Fn - mg = ma

Fn = mg

Fn = 77 x 9.8 = 754.6 N

d)

Vf = final velocity = 0 m/s

Vi = initial velocity = 1.25 m/s

During 1.80 sec : acceleration "a" is given as

a = Vf - Vi / t = 0 - 1.25 / 1.80 = - 0.694 m/s2

force equation for the man is given as

Fn - mg = ma

Fn = m (g + a) = 77 (9.8 - 0.694)

Fn = 701.16 N

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