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 ]#?

Question

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 ]#?

Amelia Adams · Accepted Answer

#Delta K= 65/6#

Writing the acceleration in terms of the force is how we start.

# 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 combine the two sides

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

As is well known, the kinetic energy is provided 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 is equal to (4/3 2^3 + 1/2 2^2+v_1)-(4/3 1^3 + 1/2 1^2+v_1), or 65/6.

Elena Albarran · Answer

#Delta K = 65/6#

Using the idea of work—which is the force integrated over the applied distance—I've determined that this is actually much simpler than the previous response.

#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 we do ultimately causes the object's kinetic energy to increase, so

#Delta K = 65/6#

Charles Aeschliman · Answer

undefined The change in kinetic energy of the object as it moves from x = 1 to x = 2 is equal to the work done by the force F(x) over that interval. Therefore, to find the change in kinetic energy, we need to calculate the work done by integrating the force function F(x) with respect to x over the interval [1, 2]. This can be expressed as:

∆KE = ∫[1 to 2] F(x) dx

Where F(x) = 4x^2 + x.

Then, ∆KE = ∫[1 to 2] (4x^2 + x) dx.

After integrating and evaluating the expression, we'll get the change in kinetic energy.