Question

you pull on a spring whose spring constant is 17 n/m, and stretch it from its equilibrium length of 0.2 m to a length of 0.3 m. estimate the work done by dividing the stretching process into two stages and using the average force you exert to calculate work done during each stage.

Answer 1

Answer:

Explanation:

We shall divide the stretch in two stage .

from .2 m to .25 m ( extension is .05 )

from .25 m to .3 m

force when length  .2 m = 0

force when length .25 m = 17 x .05 = .85 N .

average force = (0 + .85) / 2

= .425 N

work done = average force x extension

= .425 x .05

= .02125 J .

from .25 m to .3 m

force when length  .25 m = .85 N

force when length .3 m = 17 x .1 = 1.7 N .

average force = (1.7 + .85) / 2

= 1.275 N

work done = average force x extension

= 1.275 x .05

= .06375 J .