a 60 kg bicyclist going 2 m/s increased his work output by 1,800 j. what was his final velocity? m/s

Answer:

8

Explanation:

Step1: Calculate initial kinetic energy

$K_{i}=\frac{1}{2}mv_{i}^{2}=\frac{1}{2}\times60\times2^{2}=120\ J$

Step2: Calculate final kinetic energy

$K_{f}=K_{i}+W = 120 + 1800=1920\ J$

Step3: Solve for final velocity

Since $K_{f}=\frac{1}{2}mv_{f}^{2}$, then $v_{f}=\sqrt{\frac{2K_{f}}{m}}=\sqrt{\frac{2\times1920}{60}}=\sqrt{64}=8\ m/s$