學生上來請問恩師~有關彈簧的簡諧運動問題(原題目如下)

A block of mass 1.80kg is attached to a horizontal spring that has a force constant of 1.00*10^3 N/m. The spring is compressed 2.00cm and is then released from rest.

(a)Calculates the speed of the block as it passed through the equilibrium position x=0 if the surface is frictionless.

(b)Calculates the speed of the block as it passed through the equilibrium position if the coefficient of kinetic friction is 0.200.

(c) Calculates the speed of the block as it passed through the position x=-1 cm if a constant friction 4.00N retards the block`s motion from the moment it is released.

我只解了(a)F=k△x=10^3*(2*10^-2)=20N

又Fc=m*(v^2/R)→20=1.8*(v^2/2*10^-2)→v=[(2)^1/2]/3

而(b)與(c)多了摩擦力,不知其向心力是否改變(若不改變,水平分力與f...有所迷惑,若有改變,達平衡點時向心力又為何)