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