Asked by john
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.
Find (a) the amplitude of the motion.
(b) the maximum acceleration of the block.
(c) the maximum force the spring exerts on the block.
Answers
Answered by
drwls
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
= (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
Answered by
john
so is part a .162m
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