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.

To find the amplitude of the motion, maximum acceleration, and maximum force exerted by the spring on the block, we can utilize the principles of simple harmonic motion (SHM) and Hooke's Law.

(a) The amplitude of the motion represents the maximum displacement from the equilibrium position. For an object undergoing SHM with a given displacement, we can use the formula:

x = A * sin(ωt)

Where:
x is the displacement,
A is the amplitude of motion,
ω is the angular frequency of the motion, and
t is time.

Given that the displacement of the block is 0.270 m, we can find the amplitude as follows:

0.270 = A * sin(ωt)

To find A, divide both sides of the equation by sin(ωt):

A = 0.270 / sin(ωt)

(b) The maximum acceleration of the block can be determined by the relationship between acceleration and displacement in SHM:

a = -ω^2 * x

Where:
a is acceleration,
ω is the angular frequency, and
x is the displacement.

Given that the block has a displacement of 0.270 m, we can calculate the maximum acceleration using this formula.

(c) The maximum force exerted by the spring can be determined using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position:

F = -k * x

Where:
F is the force exerted by the spring,
k is the force constant of the spring, and
x is the displacement.

Given a force constant of 315 N/m and a displacement of 0.270 m, we can calculate the maximum force exerted by the spring using this formula.

Please note that to calculate ω (angular frequency), we need additional information such as the mass of the block or the period of motion.