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 ]#?

1 Answer
Feb 8, 2018

Answer:

#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#