The force applied against an object moving horizontally on a linear path is described by #F(x)= 4x^2+x #. By how much does the object's kinetic energy change as the object moves from # x in [ 1, 2 ]#?

2 Answers
Mar 11, 2016

Answer:

#Delta K= 65/6#

Explanation:

We begin by writing the acceleration in terms of the force,

# m*a=F#

#m ddot x=4x^2+x#

Multiply both sides by #dot x#

#m ddot x dot x = 4x^2 dot x + x dot x#

then integrate both sides

#1/2 m dot x ^2=4/3 x^3 + 1/2 x^2+v_1#

We know that the kinetic energy is given by

#K=1/2 m dot x^2 = 4/3 x^3 + 1/2 x^2+v_1#

The change in kinetic energy will be the difference at the two point in #x#

#Delta K=(4/3 2^3 + 1/2 2^2+v_1)-(4/3 1^3 + 1/2 1^2+v_1) = 65/6#

Mar 11, 2016

Answer:

#Delta K = 65/6#

Explanation:

I've decided that this is actually much easier than the previous answer. Let's use the concept of work which is the force integrated over the distance that is applied.

#W = int_1^2 F dx = int_1^2 (4x^2+x) dx = [(4x^3)/3+x^2/2]_1^2#
#" " = (4*2^3)/3+2^2/2-(4*1^3)/3-1^2/2 = 65/6#

The work that we do ends up increasing the kinetic energy of the object, so

#Delta K = 65/6#