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

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

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