# A 0.300 kg block is placed on a light vertical spring (k = 5.00 \times 10^3 N/m) and pushed...

## Question:

A {eq}\displaystyle 0.300 {/eq} kg block is placed on a light vertical spring {eq}\displaystyle (k = 5.00 \times 10^3 N/m) {/eq} and pushed downward, compressing the spring {eq}\displaystyle 0.100 {/eq} m. After the block is released, it leaves the spring and continues to travel upward.

What height above the point of release will the block reach if air resistance is negligible?

## Spring:

Springs are made up of high strength alloys with high elastic properties. When an external load is applied to the spring, it deforms and the ability of the spring to retain its original position after the removal of the external load is called the elasticity of the spring. Different type of springs used in engineering applications are:

1) Compression spring

2) Extension spring

3) Torsion spring, etc.

## Answer and Explanation:

We are given the following data:

- Mass of the block {eq}m=0.300\ \text{kg} {/eq}

- Stiffness of the spring {eq}k=5.00\times10^{3}\ \text{N/m} {/eq}

- Compression of the spring {eq}x=0.100\ \text{m} {/eq}

**First of all, calculating the energy stored in the spring when it is compressed by the block by using the following relation:**

$$\begin{align} E&=\dfrac{1}{2}kx^{2}\\[0.3 cm] &=\dfrac{1}{2}\times5.00\times10^{3}\ \left (\text{N/m} \right )\times(0.100\ \text{m})^{2}\\[0.3 cm] &=2.5\ \text{N.m} \end{align} $$

When the block is removed from the spring, the energy stored in the spring pushes the block in the upward direction, the height {eq}(h) {/eq} above the point of release is calculated by using the following relation:

$$\begin{align} \text{Energy stored in spring}&=\text{Potental energy gained by the block}\\[0.3 cm] E&=mgh&\left [ g\ \text{is the gravitational acceleration=9.80 m/s}^{2} \right ]\\[0.3 cm] 2.5&=0.300\ \text{kg}\times9.80\ \text{m/s}^{2}\times h\\[0.3 cm] h&=\dfrac{2.5}{0.300\times9.80}\\[0.3 cm] h&\approx \boxed{\color{blue}{.85\ \text{m}}} \end{align} $$

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