A 81 kg block is released at a 3.8 m height. the track is frictionless. the block travels down the track, hits a massless spring constant k=1888. calculate the average force

Since the track is friction less, block’s kinetic energy at the bottom of the track is equal to its potential energy at the top of the track.

PE = 81 * 9.8 * 3.8 = 3016.44 J

Work = 1/2 * 1888 * d^2

PE = Kinetic energy at the base.

1/2 * 1888 * d^2 = 3016.44

d = 1.78 approx 1.8

F = Ke = 1888 * 1.8 = 3398.4N

Average force = work / displacement

Work = change in mechanical energy

Change in mechanical energy from the release until the block is stopped due to the force of the spring = m*g*h - 0 = 81 kg * 9.8 m/s^2 * 3.8 m = 3016.44 J.

displacement of the srping:

Change in elastic potential energy = (1/2) k*x^2

=> 3016.44 J = (1/2) 1888 N/m * x^2 = > x^2 = 2 * 3016.44 J / 1888 N/m = 3.2 m^2

=> x = √3.2 m^2 = 1.79 m

=> Average force = work / displacemet = 3016.44 J / 1.79 m = 2394 N

Answer: 1,685 N