work done by a variable force

[soccerball3211]

A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. The force is given by F(x) = (1.3 - x2)(i hat) N, where x is in meters and the initial position of the block is x = 0. (a) What is the kinetic energy of the block as it passes through x = 1.5 m? (b) What is the maximum kinetic energy of the block between x = 0 and x = 1.5 m?

For part A.) I just took the Integral of the force from x=0 to x=5 and evaluated the expression. I ended up with an answer of .825J.

b.) Im a little confused with part b. Can someone help get me started?

Thanks again

 

[soccerball3211]

Correction I took the integral from 0 to 1.5 for part a!

 

[skeeter]

\(\displaystyle \L K = \frac{1}{2}mv^2\)

\(\displaystyle \L \frac{d}{dx}[K = \frac{1}{2}mv^2]\)

\(\displaystyle \L \frac{dK}{dx} = mv \frac{dv}{dx}\)

since \(\displaystyle \L v = \frac{dx}{dt}\) ...

\(\displaystyle \L \frac{dK}{dx} = m \frac{dx}{dt} \frac{dv}{dx}\)

\(\displaystyle \L \frac{dK}{dx} = m \frac{dv}{dt} = ma = F_{net} = 1.3 - x^2\)

1.3 - x² = 0 at x = 1.14 m where K = 0.988 J, a maximum.