The force applied against an object moving horizontally on a linear path is described by #F(x)=xe^x+ 1/x #. By how much does the object's kinetic energy change as the object moves from # x in [ 1, 3 ]#?

Question

The force applied against an object moving horizontally on a linear path is described by #F(x)=xe^x+  1/x #. By how much does the object's kinetic energy change as the object moves from # x in [ 1,  3 ]#?

Amelia Adams · Accepted Answer

Approximately: #41.270 J# The work done will be the integral of the force over the distance. #W = int_1^3 F(x) dx= int_1^3 (xe^x+ 1/x) dx# We'll leave the process of solving this integral to the student. The answer is: #W = 2*e^3 + log(3) = 41.270 J# Since we know that the work done is equal to the change in kinetic energy, we can say that: #Delta KE = W = 41.270 J#

Elena Albarran · Answer

undefined To find the change in kinetic energy as the object moves from $ x = 1 $ to $ x = 3 $, you can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

The work done by the force $ F(x) $ over the interval $ [1, 3] $ is given by the definite integral of $ F(x) $ with respect to $ x $ over that interval:

$$ \text{Work} = \int_{1}^{3} F(x) \, dx $$

Then, the change in kinetic energy is equal to the work done:

$$ \Delta KE = \text{Work} $$

$$ \Delta KE = \int_{1}^{3} (xe^x + \frac{1}{x}) \, dx $$

$$ \Delta KE = \left[ \frac{xe^x}{2} + \ln|x| \right]_{1}^{3} $$

$$ \Delta KE ≈ 14.812 \, \text{J} $$