If a baseball's velocity is increased to six times its original velocity, by what factor does its kinetic energy increase?

1. Break Down the Problem

To determine the increase in kinetic energy (KE) when the velocity of a baseball is increased to six times its original velocity, we start by recalling the formula for kinetic energy: $KE = \frac{1}{2} mv^2$ where $m$ is the mass and $v$ is the velocity of the object.

2. Relevant Concepts

Let:

• $v_0$ = original velocity
• $v_1 = 6v_0$ = new velocity

The kinetic energy at the original velocity ($KE_0$) is: $KE_0 = \frac{1}{2} m v_0^2$

The kinetic energy at the increased velocity ($KE_1$) is: $KE_1 = \frac{1}{2} m (6v_0)^2$

3. Analysis and Detail

Calculating $KE_1$: $KE_1 = \frac{1}{2} m (36 v_0^2) = 18 mv_0^2$

Now, we can express $KE_1$ in terms of $KE_0$: $KE_0 = \frac{1}{2} mv_0^2$ So, $KE_1 = 36 \left(\frac{1}{2} mv_0^2\right) = 36 KE_0$

4. Verify and Summarize

The increase in kinetic energy when the velocity is increased to six times the original is therefore a factor of 36.

Final Answer

The kinetic energy increases by a factor of 36.