$\begin{array}{1 1}100J\\250J\\140J\\150J\end{array} $

Answer : 100J

We first determine the spring constant,$k$.Because the force is 25N when $x$ is 0.25m,we can use Hooke's law to determine $k$

$f(x) =kx$

$25N=k\big(\large\frac{1}{4}$$m\big)$

$\therefore k =100\large\frac{N}{m}$

$f(x) =100x$

The work done by stretching the spring from a length of 1.5m to 2.5m:

$W=\int\limits_a^b f(x) dx$

$\;\;\;\;=\int\limits_{0.5}^{1.5} 100xdx$

$\;\;\;\;=\large\frac{100}{2}$$x^2\bigg|_{0.5}^{1.5}$

$\;\;\;\;=50(1.5)^2-50(0.5)^2$

$\;\;\;\;=100Nm$

$\;\;\;\;=100J$

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