Question

kinetic energy can be modeled by the formula $ke=\frac{1}{2}mv^{2}$, where m is the mass of an object and v is the velocity of an object. if all of the potential energy of the ball is transformed into kinetic energy, what is the velocity of the ball at the point of maximum kinetic energy? a 5 m/s b 10 m/s c 100 m/s d 1,250 m/s

Explanation:

Step1: Recall conservation of energy

Let the potential energy be $PE = mgh$ and kinetic energy $KE=\frac{1}{2}mv^{2}$. At the point of maximum kinetic - energy, all potential energy is converted to kinetic energy, so $mgh=\frac{1}{2}mv^{2}$.

Step2: Solve for velocity

We can cancel out the mass $m$ from both sides of the equation $mgh=\frac{1}{2}mv^{2}$, getting $gh = \frac{1}{2}v^{2}$. Assume the height $h = 5m$ and $g = 10m/s^{2}$ (standard value of gravitational acceleration near the Earth's surface). Then $10\times5=\frac{1}{2}v^{2}$, so $50=\frac{1}{2}v^{2}$. Multiply both sides by 2: $v^{2}=100$. Take the square - root of both sides, $v = 10m/s$.

Answer:

B. 10 m/s