- #1

suffian

I reasoned as follows:A particle of massmstarts from rest at timet= 0 and is moved along the x-axis with constant accelerationafrom x=0 to x=h against a variable force of magnitude F(t)=t^{2}. Find the work done.

Calculus and Analytic Geometry by George. B Thomas

The force I need to apply at any given time

*t*to maintain constant acceleration would need to be

*t*

^{2}to cancel out the force acting against me plus

*ma*to maintain a constant forward acceleration. In other words F(

*t*) =

*t*

^{2}+

*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*) =

*at*

^{2}/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 + h

^{2}/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

*t*

^{2}). But the book doesn't ever consider the acceleration.

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