Question

A 2.30 kg frictionless block is attached to an ideal spring with force constant 314 N/m . Initially the block has velocity -3.50 m/s and displacement 0.240 m .

Find the amplitude of the motion.?

Find the maximum acceleration of the block.?

Find the maximum force the spring exerts on the block.?

Answer 1

Here ,

m = 2.3 Kg

k = 314 N/m

v = 3.5 m/s and x = 0.240 m

let the amplitude of A

Using conservation of energy

0.50 * 314 * 0.240^2 + 0.50 * 2.3 * 3.5^2 = 0.50 * 314 * A^2

solving for A

A = 0.383 m

the amplitude of motion is 0.383 m

maximum acceleration is a

2.3 * a = k * A

2.3 * a = 314 * 0.383

a = 52.2 m/s^2

the maximum acceleration of the block is 52.2 m/s^2

maximum force = k * x

maximum force = 314 * 0.383

maximum force = 120.3 N

the maximum force is 120.3 N