Consider a 75.0-kg man standing on a spring scale in an elevator. Starting from

Question

Consider a 75.0-kg man standing on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.78 m/s in 0.750 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 2.00 s and comes to rest. (a) What does the spring scale register before the elevator starts to move? (b) What does the spring scale register during the first 0.750 s? (c) What does the spring scale register while the elevator is traveling at constant speed? (d) What does the spring scale register during the time it is slowing down?

Explanation / Answer

a)weight registered by spring scale=mg=75*9.8=735 N (ans)

b)accleration,a=v/t=1.78/0.75=2.37 m/sec^2

weight registered by spring scale=mg +ma =912.75 N (ans)

c)weight registered by spring scale=mg=735 N (ans)

d)v=u -a't =0

a'=1.78/2=0.89 m/sec^2

so weight registered by spring scale=mg -ma'=668.25 N (ans)

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