Calculus problem

[Daisydukes]

Problem:


"The spring of a spring balance is 5.0in long when there is no load on the balance, and it is 9.6in long with 8ft/lbs hung from it. How many ft/lbs is done in stretching the spring from 5.0in to 12.2in?"

So far I think I found the constant of the spring which is 8=9.6k which = 0.83333
and I know the integral will be from 5.0 to 12.2 "But I am stumped as to what to do next....thanks for the help <3 ;-)

 

[Deleted member 4993]

Problem:


"The spring of a spring balance is 5.0in long when there is no load on the balance, and it is 9.6in long with 8ft/lbs hung from it. How many ft/lbs is done in stretching the spring from 5.0in to 12.2in?"

So far I think I found the constant of the spring which is 8=9.6k which = 0.83333
and I know the integral will be from 5.0 to 12.2 "But I am stumped as to what to do next....thanks for the help <3 ;-)

Please check your problem statement for accuracy.

I am not aware of any common physical entity with that dimension.

 

[HallsofIvy]

DaisyDukes clearly meant "ft-lbs" rather than "ft/lbs".

DaisyDukes, for constant force, "work" is "force times distance". If we have a variable force, we can imagine the distance cut into many small intervals, of length "dx", and take the force to be constant, equal to f(x) for some x in each interval. Then the work done in each interval is approximately f(x)dx and the total work is approximately \(\displaystyle \sum f(x) dx\). To make that exact, we take the limit as the length of each interval goes to 0 and the number of intervals goes to infinity: in other words it becomes the integral \(\displaystyle \int_a^b f(x)dx\).

If you were given a problem like this I would be very surprised if you had not already seen something like that in class.