# A 0.400 kg mass is attached to a horizontal spring with a spring constant of 50.0 N/m. The mass...

## Question:

A 0.400 kg mass is attached to a horizontal spring with a spring constant of 50.0 N/m. The mass is initially stretched a distance of 48.0 cm from its equilibrium position. The mass is then released from rest and slides along a horizontal table. A) Assume the table is frictionless and there is no air resistance. Select the energy types which change from the time the mass is released to the time it is compressed a distance of 39.0 cm on the other side of equilibrium. K , Ug, Uel, Uint B) What will be the speed of the mass when it is compressed to the distance of 39.0 cm from equilibrium? C) Due to the kinetic friction from the table, after being released as before (from rest, stretched a distance of 48.0 cm from equilibrium), the mass reaches its maximum compression 40.0 cm on the other side of equilibrium. Find the change in the mechanical energy of the mass-spring from when it is released to when it reaches its maximum compression. D) Using your answer to part C, find the work done by the external force of the table on the mass. E) Use your answer to part D to solve for the coefficient of kinetic friction between the table and the mass. F) In this system, select which types of energy must change from the initial state (released from rest stretched 48 cm from equilibrium) to the final state (maximum compression 40 cm from equilibrium)? Ug, Uel, Uint, K G) Use conservation of energy to solve for the change in the internal energy of the system.

## Energy Conservation:

In science, the law of conservation of energy expresses that the complete vitality of a separated framework stays steady; it is said to be moderated after some time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.

a) Initially, the block will have only potential energy. Now when the block is released the potential energy is converted to kinetic energy. When the block reaches the equilibrium, kinetic energy decreases at that point so it has some potential energy and some kinetic energy

b)

Using conservation of energy

{eq}PE_i+K.E_i =P.E_f +K.E _f {/eq}

{eq}\frac{kx_i^2}{2}+0 =\frac{k x_f^2}{2} +\frac{mv^2}{2} {/eq}

{eq}v = \sqrt\frac{k(x_i^2-x_f^2)}{m} {/eq}

{eq}v = \sqrt\frac{50(0.48^2-0.39^2)}{0.400} {/eq}

= 3.128 m/sec

c)

Change in mechanical energy is

{eq}E = \frac{kx_i^2}{2}-\frac{kx_f^2}{2} {/eq}

{eq}E = \frac{k(x_i^2-x_f^2)}{2} {/eq}

{eq}E = \frac{50(0.48^2-0.40^2)}{2} {/eq}

=1.76 J

d)

The work done by the external forces is equal to change in mechanical energy

W = 1.76 J