Question

An elevator car of mass 805 kg falls from rest 4.00 m, hits a buffer spring, and then travels an additional 0.500 m, as it compresses the spring by a maximum of 0.500 m. What is the force constant of the spring

Answer 1

Answer:

2665.6 N/m

Explanation:

The potential energy of elevator will convert to PE of the spring

Here,

m=8.5 kg

h=4.0 m

g=9.8m/s2

PE of elevator =mgh = 8.5×9.8×4.0=333.2 J

PE of spring = 1/2 k $x^{2}$

where,

k=? (to be found)

x= 0.5 m (distance by compressing the spring)

==> PE of spring = 1/2× k× 0.5×0.5 = 0.125k

Since .

PE of elevator = PE of the spring

==> 333.2 =0.125k

==> k = 333.2÷0.125=2665.6 N/m