Spring Force (hooke's law)

Question

A spring scale hangs from the ceiling of an elevator. Attached to the scale is a 25N weight.

A)Draw an FBD of the weight and establish Fnet=ma in the vertical direction, solving for Fs.

B) What is the scale reading when the elevator is still? show all work

C) What is the scale reading when the elevator is accelerating up at 1.5 m/s^2?

D) How far does the spring compress/stretch in each situation if k=12N/cm?

Answer 1

FBD has Fs vertical up and mg vertical down. Let up be the positive direction. Fs=m(a+g) a)

b) Elevator is still then a=0 thus Fs=mg=25*9.81=245.3 N

c) Elevator accelerating up means a=+1.5 m/s² thus Fs=25*(1.5+9.81)=282.8 N

d)dx=Fs/K ... where K=12N/cm dx in c) = 23.6 cm, in b)=20.4cm