Determining Work and Fluid Force

Key Questions

  • How mush work is done in lifting a 40 kilogram weight to a height of 1.5 meters?

    The answer is #588J#.

    Always start with the definition:

    #W=int_a^b F(x)dx#

    Sometimes, the simple problems are the hardest because it looks too easy so we tend to add unnecessary things. Since this is a vertical lift, we are dealing with gravity which is 9.8 #m/s^2#. So,

    #F(x)=9.8m/(s^2)*40kg=392N#

    Remember that work is force times distance. We have force which is just a constant function. And we have distance which is #dx#. The tendency is to add an #x# into #F(x)=392N#, but that would be incorrect. Now, let's put it together:

    #a=0#
    #b=1.5#
    #W=int_0^(1.5) 392 dx#
    #=392x|_0^(1.5)#
    #=588J#

    Amory W. · · Aug 21 2014
  • If the work required to stretch a spring 1 foot beyond its natural length is 12 foot-pounds, how much work is needed to stretch it 9 inches beyond its natural length?

    The answer is #(27)/4# ft-lbs.

    Let's look at the integral for work (for springs):

    #W=int_a^b kx \ dx = k \ int_a^b x \ dx #

    Here's what we know:

    #W=12#
    #a=0#
    #b=1#

    So, let's substitute these in:

    #12=k[(x^2)/2]_0^1#
    #12=k(1/2-0)#
    #24=k#

    Now:

    9 inches = 3/4 foot = #b#

    So, let's substitute again with #k#:

    #W=int_0^(3/4) 24xdx#
    #=(24x^2)/2|_0^(3/4)#
    #=12(3/4)^2#
    #=(27)/4# ft-lbs

    Always set up the problem with what you know, in this case, the integral formula for work and springs. Generally, you will need to solve for #k#, that's why #2# different lengths are provided. In the case where you are given a single length, you're probably just asked to solve for #k#.

    If you are given a problem in metric, be careful if you are given mass to stretch or compress the spring vertically because mass is not force. You will have to multiply by 9.8 #ms^(-2)# to compute the force (in newtons).

    Amory W. · 22 · Aug 21 2014
  • How do you determine the amount of work needed for movement of objects?

    If #F(x)# denotes the amount of force applied at position #x# and it moves from #x=a# to #x=b#, then the work #W# can be found by

    #W=int_a^b F(x)dx#.

    I hope that this was helpful.

    Wataru · 1 · Oct 31 2014

Questions

Applications of Definite Integrals