Question

select the equation that shows that a 600000 - j change in kinetic energy occurs for an airplane that is moved 300 m with a force of 2000 n.

$w=delta ke = fd^{2}=(2000 n)(300 m)^{2}=600000 j$
$w=delta ke = f/d=(2000 n)/(300 m)=600000 j$
$w=delta ke = fd=(2000 n)(300 m)=600000 j$
$w=delta ke = 2fd=2(2000 n)(300 m)=600000 j$

Explanation:

Step1: Recall work - energy theorem

The work - energy theorem states that the work done on an object is equal to the change in its kinetic energy, i.e., $W=\Delta KE$. The work done by a constant force $F$ over a displacement $d$ is given by the formula $W = Fd$.

Step2: Substitute given values

We are given that $F = 2000\ N$ and $d=300\ m$. Substituting these values into the work formula $W = Fd$, we get $W=(2000\ N)\times(300\ m)=600000\ J$.

Answer:

$W=\Delta KE = Fd=(2000\ N)(300\ m)=600000\ J$ (the third option)