Question

A spring of spring constant k =4×10^3N/m is compressed by 2 cm then determine potential energy stored in spring

Answer 1

Answer:

• Potential Energy (E) of the Spring is 40 Joules.

Given:

1. Spring Constant (K) = 4 × 10³ N/m.
2. Compressed By (x) = 2 cm.

Explanation:

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Spring Potential Energy:-

• It's an Energy which gets Stored in the Spring while it's Compressed (or) Elongation.
• It has the Same units As that of Work done I.e. Joules.

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From the Expression we Know,

$\large\bigstar \: {\boxed{\tt E = \dfrac{1}{2}Kx^2}}$

$\bold{Here}\begin{cases}\text{E Denotes Potential Energy} \\ \text{K Denotes Spring constant} \\ \text{x Denotes the Compression}\end{cases}$

Now,

$\large{\boxed{\tt E = \dfrac{1}{2} Kx^2}}$

Substituting the values,

$\large{\tt \longmapsto E = \dfrac{1}{2} \times 4 \times 10^3 \: N/m \times 2 \: cm}$

As the Distance of Compression is in Cm we Need to convert it into Meters.

$\large{\tt \longmapsto E = \dfrac{1}{2} \times 4 \times 10^3 \: N/m \times 2 \times 10^{-2} \: m}$

∵[1 cm = 10⁻² m]

$\large{\tt \longmapsto E = \dfrac{1}{2} \times 4 \times 10^3 \times 2 \times 10^{-2}}$

$\large{\tt \longmapsto E = \dfrac{4}{2} \times 2 \times 10^3 \times 10^{-2}}$

$\large{\tt \longmapsto E = \cancel{\dfrac{4}{2}} \times 2 \times 10^3 \times 10^{-2}}$

$\large{\tt \longmapsto E = 2 \times 2 \times 10^3 \times 10^{-2}}$

$\large{\tt \longmapsto E = 2 \times 2 \times 10^{+ 3 - 2}}$

$\large{\tt \longmapsto E = 4 \times 10^{1}}$

$\large{\tt \longmapsto E = 4 \times 10}$

$\huge{\boxed{\boxed{\tt E = 40 \: J}}}$

∴ Potential Energy (E) Stored in the Spring is 40 Joules.

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