Ablock of mass 1.6 kg is attached to a horizontal spring that has a spring constant of 1,000 latex: \frac{n}{m}n m. the spring is compressed 2 cm and is then released from rest. calculate the speed of the block as it passes through the equilibrium position if a constant friction force of 4 n acts on the block from the moment it is released.

3.52 m/s

Explanation:

work done by the compressing the spring = 1/2 K e² where K is the force constant = 1000 N/m, e is the compression = 2cm = (2 / 100) to convert it to m we divide by 100  = 0.02 m

work done by compressing the spring = elastic potential energy stored in the spring = 0.5 × 1000 × 0.02 = 10 J

work done by force of friction to hinder the motion = F × d = 4 × 0.02 m = 0.08 J

Kinetic energy of the body = work done by compressing the spring - work done by force of friction against the motion = 10 - 0.08 = 9.92

9.92 = 1/2 m v² where  m is the mass of the body which = 1.6 kg and v is the speed as it passes through the equilibrium point

9.92 = 1/2 × 1.6 × v²

9.92 × 2 / 1.6 = v²

v² = 19.84 / 1.6 = 12.4

v = √12.4 = 3.52 m/s

prolly a beacuse the momentuem the car has increases while going down hill