Physics HW Help?
A 1600.0 g mass is on a horizontal surface with μk = 0.34, and is in contact with a massless spring with a force constant of 655.0 N/m which is compressed. When the spring is released, it does 1.18 J of work on the mass while returning to its equilibrium position. Calculate the distance the spring was compressed.
What is the velocity of the mass as it loses contact with the spring?
- JimLv 74 years agoFavorite Answer
SPE = 1.18 = kx² = (655.0)x²
x² = 1.18/655
x = 0.0424 m ANS-1
weight of mass = mg = (1.600)(9.81) = 15.696 N
kinetic friction force against mass movement = µk(15.696) = 0.34(15.696) ≈ 5.34 N
work done by kinetic friction force = 5.34(0.0424) = 0.226 J
KE of mass as it leaves spring = 1.18 - 0.226 = 0.954 J
1/2mV² = 0.954
V² = 1.19216
V ≈ 1.09 m/s ANS-2