A 91 kg block is released at a 6 m height as shown. The track is frictionless. The block travels down the track, hits a spring of force constant k = 1968 N/m. The acceleration of gravity is 9.8 m/s

Determine the compression of the spring x from its equilibrium position before coming to rest momentarily. Answer in units of m.

h=6m

m=91kg

k=1968N/m

Question

Thomas Waller

Physics

6 May, 17:31

A 91 kg block is released at a 6 m height as shown. The track is frictionless. The block travels down the track, hits a spring of force constant k = 1968 N/m. The acceleration of gravity is 9.8 m/s

Determine the compression of the spring x from its equilibrium position before coming to rest momentarily. Answer in units of m.

h=6m

m=91kg

k=1968N/m

+5

Answers (1)

Know the Answer?

Josue Peck · Answer 1

Initial gravitational potential energy = final elastic potential energy

initial gravitational potential energy = m x g x h

= 91 x 9.8 x 6 = 5351 J

initial gravitational potential energy = final elastic potential energy = 5351J

final elastic potential energy = 1/2 x k x d^2 = 1/2 x 1968 x d^2 = 984 x d^2

5351 = 984 x d^2

d = √ (5351/984)

d=2.33m