# A .20 kg object, attached to a spring with spring constant k = 1000 n/m, is moving on a...

## Question:

A .20 kg object, attached to a spring with spring constant k = 1000 n/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude of 0.080 m. What is its speed at the instant when its displacement is 0.040 m? (Hint: Use conservation of energy.)

a) 9.8 m/s

b) 4.9 m/s

c) 49 cm/s

d) 24.5 cm/s

## Spring Potential Energy:

Spring potential energy of the spring is the energy that is stored in the spring when it is compressed by applying some forces on it and higher will be the deformation of the spring higher will be the spring potential energy.

For the expression of spring potential energy :

{eq}P.E_s=\frac{1}{2}kx^2 {/eq}

Here, the spring constant is {eq}k {/eq}

Displacement of the spring is {eq}x {/eq}

Given

Mass of the object is {eq}m=20\ kg {/eq}

Spring constant is {eq}k=1000\ N/m {/eq}

The amplitude of the spring is {eq}A=0.080\ m {/eq}

The maximum displacement of the spring is {eq}x=0040\ m {/eq}

The total energy of oscillation will be equal to the spring potential energy and the kinetic energy of the block:

{eq}\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2\\ v^2=\frac{k}{m}(A^2-x^2)\\ v^2=\frac{1000}{20}(0.080^2-0,040^2)\\ v^2=0.24\\ v=0.49\ m/s v=49.1\ cm/s {/eq}

Thus, the speed of the block will be 49 cm/s and the correct option will be (c) 