# A block of mass 0.45 kg can slide over a frictionless horizontal surface. It is attached to a...

## Question:

A block of mass 0.45 kg can slide over a frictionless horizontal surface. It is attached to a spring whose stiffness constant is k = 18 N/m. The block is pulled 38 cm and let go.

a) What is its maximum speed?

b) What is its speed when the extension is 30 cm?

c) At what position is the kinetic energy equal to the potential energy?

## Conservation Of Energy :

Conservation of energy principle relates the different type of energy during the conversion from one form of energy to the other and according to which the energy cannot be created or destroyed during the energy conservation, for example, the potential energy is converted into the kinetic energy when the object is allowed to fall from certain height.

## Answer and Explanation:

Given

Mass of the block is {eq}m= 0.45\ kg {/eq}

Spring constant of the spring is {eq}k= 18\ N/m {/eq}

Distance by which the spring is stretched is {eq}s= 38\ cm {/eq}

Now for the conservation of energy spring potential energy will be equal to the kinetic energy of the block :

{eq}\frac{1}{2}kx^2=\frac{1}{2}mv^2\\ \frac{1}{2}(18)(0.38)^2=\frac{1}{2}(0.45)\times v^2\\ 1.299=0.225\ v^2\\ v=2.40\ m/s {/eq}

Thus, the maximum speed of the block is 2.40 m/s

(b)

Now for the speed when the extension is 30 cm:

{eq}\frac{1}{2}kx^2=\frac{1}{2}mv^2\\ \frac{1}{2}(18)(0.3)^2=\frac{1}{2}(0.45)\times v^2\\ 0.81=0.225\ v^2\\ v=1.89\ m/s {/eq}

Thus, the speed of the block is 1.89 m/s

(c)

Now for the position where spring potential energy equal to kinetic energy :

{eq}x=\frac{A}{2}\\ x=\frac{0.35}{2}\\ x=0.175\ m {/eq}

Thus, the position is 0.175 cm