Question

one end of a horizontal spring(k=80N/m)is held fixed while an external force is applied to the free end,stretching it slowly from×1=o to ×2=4cm.what is work done by the applied force on the spring​

Answer 1

Answer:

6.4 Joules

Explanation:

For springs, Hooke's law states that;

F = ke

where F is a force applied, e is the extension and k is the spring constant.

Work done in a spring is the same as the potential energy stored in the spring. So that;

Work done = $\frac{1}{2}$ k$e^{2}$

e = $x_{2}$ - $x_{1}$

$x_{2}$ = 4 cm = 0.4 m

$x_{1}$ = 0

So that,

$x_{2}$ - $x_{1}$ = 0.4 m

Thus,

Work done = $\frac{1}{2}$ x 80 x $(0.4)^{2}$

                   = 40 x 0.16

                   = 6.4

Work done = 6.4 J

The work done by the force on the spring is 6.4 Joules.