Question

The potential energy of an object attached to a spring is 2.10 J at a location where the kinetic energy is 1.90 J. If the amplitude ???? of the simple harmonic motion is 22.0 cm, calculate the spring constant ???? and the magnitude of the largest force ????spring, max that the object experiences.

Answer 1

Answer:

The spring constant and the magnitude of the largest force are 165.28 N/m and 36.362 N.

Explanation:

Given that,

Potential energy = 2.10 J

Kinetic energy = 1.90 J

Amplitude = 22.0 cm

We need to calculate the spring constant

$Total.Energy=\dfrac{1}{2}kA^2$

Put the value into the formula

$2.10+1.90=\dfrac{1}{2}\times k\times(22.0\times10^{-2})^2$

$k=\dfrac{4\times2}{22.0\times10^{-2}^2}$

$k=165.28\ N/m$

We need to calculate the largest force

Using restoring force

$F=-kx$

For maximum force,

$F=-kA$

Put the value into the formula

$F=-165.28\times22.0\times10^{-2}$

$F=36.362\ N$

Hence, The spring constant and the magnitude of the largest force are 165.28 N/m and 36.362 N.

Answer 2

(a) The spring constant is 165.29 N/m.

(b) The magnitude of the largest force of the spring is 36.36 N.

Conservation of energy

The total energy of the spring is determined by applying the principle of onservation of energy as shown below;

P.E + K.E = U

2.1 + 1.9 = ¹/₂kA²

4 = ¹/₂ x k x (0.22)²

k = 165.29 N/m

Mangitude of the largest force

The magnitude of the largest force of the spring is calculated as follows;

F = kx

F = 165.29 x 0.22

F = 36.36 N

Learn more about conservation of energy here: https://brainly.com/question/166559