A. elastic P.E. = (1/2) k Xa^2
Xa must be in meters, since k has units of N/m.
B. The difference in elastic P.E at the two end positions is the work done against friction.
(1/2)k [Xa^2 - Xc^2) = M g ì (Xa + Xc)
Solve for Xc
A mass of 0.755 kg is attached to a horizontal spring of constant 403 N/m as shown. The system is compressed a distance xa = 20.8 cm from its equilibrium position and released from rest. The coefficient of friction on the surface is μ = 0.354. The mass comes to rest again at some distance xc on the other side of the equilibrium position.
A. What is the elastic potential energy at point A?
B. What is the deformation of the spring at point C?
1 answer
1 answer