Question

Consider the mass-on-a-spring system as shown in the figure below. The spring has a spring constant of 1.81e+3 N/m, and the block has a mass of 0.988 kg. There is a constant force of kinetic friction between the mass and the floor of 1.79 N. Starting with the spring compressed by 0.172 m from its equilibrium position, how far will the block travel once it leaves the spring? (Assume that block leaves the spring at at the spring's equilibrium position, marked x=0 in the figure.

Answer 1

K1Answer:

s= 6.5cm

Step-by-step explanation:

Point 1: just right before the block leaves the spring

Point 2: the block has v2=0 (stops moving)

Apply Kinetic-Work Theorem:

K1 + U1 + Wother = K2 + U2

K1=0

U1= 1/2×k×x² (k= 1.81e+3 N/m and x= -0.172 m)

Wother = F×s= 1.79×s

K2=0

U2=0

=> s= 6.5cm

This is my attempt to solve. Let me know if this isn't right