Question

A 2.40 kg frictionless block is attached to an ideal spring with force constant 317 N/m . Initially the block has velocity -3.61 m/s and displacement 0.210 m .

Part A

Find the amplitude of the motion.

Part B

Find the maximum acceleration of the block.

Part C

Find the maximum force the spring exerts on the block.

Answer 1

Here,

Amplitude is A

m = 2.4 Kg

k = 317 N/m

part A)

Using conservation of energy

0.50 * m * v^2 + 0.50 * k * x^2 = 0.50 * k * A^2

0.50 * 2.4 * 3.61^2 + 0.50 * 317 * 0.21^2 = 0.50 * 317 * A^2

solving for A

A = 0.38 m

the amplitude is 0.38 m

part B)

k * A = m * a

317 * 0.38 = 2.4 * a

a = 50.2 m/s^2

the maximum acceleration of the block is 50.2 m/s^2

part C)

maximum force = m * a

maximum force = 2.4 * 50.2

maximum force = 120.5 N