Question

calculate work done to pull a spring




​

Answer 1

To find :

Work done in pulling a spring ?

Solution:

Let's assume that a spring has been pulled by distance of "x" from its mean, unstretched position.

So, the magnitude of spring force at that stretched position :

$\therefore \: F =- kx$

$\implies \: |F| = kx$

• Since the operator is providing a force which is overcoming the spring force , work will be positive as the applied force is in the same direction as the displacement.

"k" is spring constant.

Now, the net work done :

$\therefore \: dW = F \times dx$

$\implies\: dW = (kx) \times dx$

Integrating on both sides:

$\displaystyle \implies\: \int_{0}^{W} dW = k \int_{0}^{x} x \times dx$

$\displaystyle \implies\: W = k \times \dfrac{ {x}^{2} }{2}$

$\displaystyle \implies\: W = \dfrac{1}{2} k{x}^{2}$

• I recommend you to always calculate the work done (while pulling a spring) from the mean position (where the spring is relaxed and unstretched).

• It prevents any confusion.

So, net work done is:

$\boxed{ \bold{\: W = \dfrac{1}{2} k{x}^{2}}}$

Answer 2

To derive:-

Formula for work done to pull a spring (by an operator) $(\sf W.)$

Answer:-

Let the initial position of the spring be $\sf x = x_i,$ and the final position be $\sf x = x_f.$

From Hooke's law,

$\sf Spring\;force = F_s = - kx$

But, since we have to find work done by the operator,

$\sf Force_{(Man)} = - (-kx) = kx \quad \quad \dots (1)$

Hence, work done by MAN to pull the spring will be,

$\sf dW = F\:dx$

$\longrightarrow \displaystyle \sf \int_{0}^{W} dW = \int_{x_i}^{x_f} F\:dx$

$\longrightarrow \displaystyle \sf W = \int_{x_i}^{x_f} kx\:dx\quad [From\:(1)]$

$\longrightarrow \displaystyle \sf W = \bigg[k\; \; \dfrac{x^{(1+1)}}{1+1}\bigg]_{x_i}^{x_f}$

$\longrightarrow \displaystyle \sf W = \bigg[\dfrac{1}{2}kx^2\bigg]_{x_i}^{x_f}$

$\longrightarrow \underline{\underline{\sf{\green{W = \dfrac{1}{2}k\big(x_f { }^2 - x_i { }^2 \big) }}}}$