Question

an 80 kg skateboarder moving at 3 m/s pushes off with her back foot to move faster. if her velocity increases to 5 m/s, what is her change in kinetic energy as a result?
j
how much work did she perform?
j

Explanation:

Step1: Recall Kinetic Energy Formula

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

Step2: Calculate Initial Kinetic Energy

Given $m = 80\ kg$ and $v_1 = 3\ m/s$, initial KE ($KE_1$) is $\frac{1}{2} \times 80 \times 3^2 = 40 \times 9 = 360\ J$.

Step3: Calculate Final Kinetic Energy

With $v_2 = 5\ m/s$, final KE ($KE_2$) is $\frac{1}{2} \times 80 \times 5^2 = 40 \times 25 = 1000\ J$.

Step4: Find Change in Kinetic Energy

Change in KE ($\Delta KE$) is $KE_2 - KE_1 = 1000 - 360 = 640\ J$.

Step5: Relate Work and Kinetic Energy (Work - Energy Theorem)

The work - energy theorem states that work done ($W$) is equal to the change in kinetic energy, so $W=\Delta KE = 640\ J$.

Answer:

Change in kinetic energy: $\boxed{640}$ J
Work performed: $\boxed{640}$ J