Question

The force applied against an object moving horizontally on a linear path is described by `F(x)=x^2-x+1`. By how much does the object's kinetic energy change as the object moves from `x in [ 1, 2 ]`?

Answer 1

`11/6 J`

Explanation:

Suppose,initially the object had a velocity of `u` after going from `x=1 to x=2` its velocity became `v`,so change in kinetic energy = `1/2 mv^2 - 1/2 m*u^2`

Given,

`F=x^2-x+1`

Or, `m×a=x^2-x+1`

Or, `m v(dv)/(dx) = x^2-x+1`

Or, `m v dv = x^2 dx -x dx +dx`

Or, `m int_u^v dv = int_1^2 x^2 dx - int_1^2 x dx + int_1^2 dx`

Or, `(mv^2)/2- (m*u^2)/2= 11/6`