A very light ideal spring having a spring constant (force constant) of 820 N/m is used to lift a 2.2 kg tool with an upward acceleration of 3.25 m/s2. If the spring has negligible length when it us not stretched, how long is it while it is pulling the tool upward?
given
k = 820 N/m
m = 2.2 kg
a = 3.25 m/s^2
Net force acting on the tool, Fnet = F_spring - m*g
m*a = F_spring - m*g
F_spring = m*g + m*a
= 2.2*9.8 + 2.2*3.25
= 28.71 N
let x is the extension of the spring.
now use, F_spring = k*x
==> x = F_spring/k
= 28.71/820
= 0.0350 m
so, length of the spring = x
= 0.0350 m (or) 3.50 cm <<<<<<<<<<<-----------Answer
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