Work with Variable Force

  • Thread starter suffian
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  • #1
I just did the following textbook problem (from my Calculus book under the chapter review from Applications of Integrals) but came up with an answer that differs from the one at back of the book:

A particle of mass m starts from rest at time t = 0 and is moved along the x-axis with constant acceleration a from x=0 to x=h against a variable force of magnitude F(t)=t2. Find the work done.

Calculus and Analytic Geometry by George. B Thomas
I reasoned as follows:
The force I need to apply at any given time t to maintain constant acceleration would need to be t2 to cancel out the force acting against me plus ma to maintain a constant forward acceleration. In other words F(t) = t2+ma.

If the acceleration is always a and the particle starts from rest at x=0 then
x''(t) = a
x'(t) = at
x(t) = at2/2

This means that once the particle reaches a position x, the following amount of time has elapsed:
t(x) = sqrt(2x/a), t > 0

Which in turn leads to the force that needs to applied at a given position x:
F( sqrt(2x/a) ) = 2x/a + ma

We now use the work formula W= ∫a..b F(x)dx :
W = ∫0..h (2x/a + ma)dx
W = mah + h2/a

On the otherhand, the book gives the following result:
W = 4h/3 sqrt(3mh)

I don't see how they arrived at their results since it doesn't take into account the acceleration of the particle. Surely, the faster the particle is accelerated the work should take toward a limiting value of mah(force * distance) because the particle will reach its destination faster and therefore have less resistance to deal with (remember the resistance was t2). But the book doesn't ever consider the acceleration.

Has the book made an error, or have i misinterpreted the problem?

Answers and Replies

  • #2
man, the units don t even work out right for the answer the book gave. that s a sure fire way to tell a wrong answer.

plus, i didn t see any mistakes in your work.