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

6.4 Joules

Step-by-step 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 =
`(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 =
`(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.