Question

Consider an elevator carrying Kermit the frog weighing 4000.0 N is held 5.00 m above a spring with a force constant of
8000.0 N/m. The elevator falls onto the spring while subject to a frictional force (brake) of 1000.0 N. Determine the
maximum compression distance, x, of the spring.

Answer 1

Answer:

The maximum compression distance of the spring is 0.375 m

Explanation:

The given parameters are;

The weight of the elevator and the frog = 4,000.0 N

The location of the elevator above the spring = 5.00 m

The force constant of the spring, k = 8,000.0 N/m

The frictional force of the brakes = 1,000.0 N

The net force, F, of the elevator on the spring is F = 4,000.0 N - 1,000.0 N = 3,000.0 N

F = 3,000.0 N

The maximum compression distance of the spring, x is given as follows;

F = k × x

Substituting the known values gives;

3,000.0 N = 8,000.0 N/m × x

∴ x = (3,000.0 N)/(8,000.0 N/m) = 0.375 m

x = 0.375 m = 37.5 cm

The maximum compression distance of the spring, x = 0.375 m.