Question

select the equation that shows that the kinetic energy of a 3.0 - kg book tossed across the room at a speed of 2 m/s. (1 kg(m/s)^2 = 1 j)
○ ke = \frac{1}{2}mv=\frac{1}{2}(3.0 kg)(2.0 m/s)=6.0 j
○ ke = \frac{1}{4}mv^{2}=\frac{1}{4}(3.0 kg)(2.0 m/s)^{2}=6.0 j
○ ke = \frac{1}{2}mv^{2}=\frac{1}{2}(3.0 kg)(2.0 m/s)^{2}=6.0 j
○ ke = mv^{2}=(3.0 kg)(2.0 m/s)^{2}=6.0 j

Explanation:

Step1: Recall kinetic - energy formula

The formula for kinetic energy is $KE=\frac{1}{2}mv^{2}$, where $m$ is the mass and $v$ is the velocity.

Step2: Substitute given values

Given $m = 3.0\ kg$ and $v=2.0\ m/s$. Substitute into the formula: $KE=\frac{1}{2}(3.0\ kg)(2.0\ m/s)^{2}$.
Calculate $(2.0\ m/s)^{2}=4.0\ (m/s)^{2}$, then $\frac{1}{2}(3.0\ kg)\times4.0\ (m/s)^{2}=6.0\ J$.

Answer:

C. $KE=\frac{1}{2}mv^{2}=\frac{1}{2}(3.0\ kg)(2.0\ m/s)^{2}=6.0\ J$