The force applied against an object moving horizontally on a linear path is described by #F(x)=2x +4 #. By how much does the object's kinetic energy change as the object moves from # x in [ 3, 5 ]#?

Question

The force applied against an object moving horizontally on a linear path is described by #F(x)=2x +4  #. By how much does the object's kinetic energy change as the object moves from # x in [ 3,  5    ]#?

Nora Arzate · Accepted Answer

#Delta E_k=-16 J# #W:" work doing"# #E_k:" changing of the kinetic energy"# #W=Delta E_k=int_a^b F(x) d x# #W=Delta E_k=int _3^5(2x+4)*d x# #Delta E_k=[2*1/2*x^2+4*x]_3^5# #Delta E_k=[x^2+4x]_3^5# #Delta E_k=(5^2+4*5)-(3^2+4*3)# #Delta E_k=5-21# #Delta E_k=-16 J#

Adam Arnold · Answer

undefined To find the change in kinetic energy, we need to calculate the work done by the force $F(x)$ as the object moves from $x = 3$ to $x = 5$. This can be done using the work-energy theorem, which states that the work done by the net force on an object is equal to the change in its kinetic energy.

First, we need to find the net force by taking the derivative of the given force function with respect to $x$, which gives us $F'(x) = 2$. This means the net force is constant, and we can find it by evaluating $F'(x)$ at any point within the interval. Since $F'(x)$ is constant, the work done is simply the force times the distance traveled, which is $F'(x) \times \Delta x$.

Substituting the values, we get:

$$
F'(x) = 2 \quad \text{and} \quad \Delta x = 5 - 3 = 2
$$

So, the work done by the force is $2 \times 2 = 4$ Joules.

Since work done equals the change in kinetic energy, the change in kinetic energy of the object as it moves from $x = 3$ to $x = 5$ is 4 Joules.