Question

a 2kg block is attached to a horizontal ideal spring with a spring constant of 200 Newton per minute. when the spring has its equillibrium length of block is given a speed of 5mps. what is the maximum elongation of the spring

Answer 1

Answer:

The maximum elongation of the spring is 0.5 meters.

Explanation:

The statement is incorrect. The correct form is:

A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200 newton per meter. When the spring has its equillibrium length, the block has a speed of 5 meters per second. What is the maximum elongation of the spring?

The block experiments a simple harmonic motion, where there are no non-conservative forces and the total energy is the sum of translational kinetic energy of the mass and the elastic potential energy of the spring. The maximum elongation of the spring is done when elastic potential energy reach its maximum. By the Principle of Energy Conservation, the maximum elastic potential energy is equal to the maximum translational kinetic energy, which corresponds to the instant when the mass reaches the equilibrium position. Then, the equation modelling the system is:

$U_{max} = K_{max}$ (1)

Where:

$U_{max}$ - Maximum elastic potential energy of the spring, measured in joules.

$K_{max}$ - Maximum translational kinetic energy of the mass, measured in joules.

By definitions of the maximum elastic potential energy of the spring and the maximum translational kinetic energy of the mass, the expression above is expanded and simplified:

$\frac{1}{2}\cdot k\cdot x_{max}^{2} = \frac{1}{2}\cdot m \cdot v_{max}^{2}$

$x_{max} = \sqrt{\frac{m}{k} }\cdot v_{max}$ (2)

Where:

$x_{max}$ - Maximum elongation of the spring, measured in meters.

$m$ - Mass, measured in kilograms.

$k$ - Spring constant, measured in newtons per meter.

$v_{max}$ - Maximum speed of the mass, measured in meters per second.

If we know that $m = 2\,kg$, $k = 200\,\frac{N}{m}$ and $v_{max} = 5\,\frac{m}{s}$, then the maximum elongation of the spring is:

$x_{max} = \sqrt{\frac{2\,kg}{200\,\frac{N}{m} } }\cdot \left(5\,\frac{m}{s} \right)$

$x_{max} = 0.5\,m$

The maximum elongation of the spring is 0.5 meters.