A 2.10 kg frictionless block is attached to an ideal spring with force constant 315N/m . Initially the block has velocity -3.75m/s and displacement 0.270m .

Question

A 2.10 kg frictionless block is attached to an ideal spring with force constant 315N/m . Initially the block has velocity -3.75m/s  and displacement 0.270m .
Find (a) the amplitude of the motion.
 (b) the maximum acceleration of the block.
 (c) the maximum force the spring exerts on the block.

drwls · Accepted Answer

Total mechanical energy
= (M/2)V^2 + (k/2)X^2 = constant
= 14.06 + 11.48 = 25.54 J
(using the given displacement and velocity data)

k = 315 N/m

(a) For the amplitude (Xmax), set V = 0 and solve
(k/2)X^2 = 25.54

for X

(b) maximum acceleration = k*Xmax/M
(c) k*Xmax

john · Answer

so is part a .162m