Question

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

Explanation:

Step1: Recall Kinetic Energy Formula

The kinetic energy formula is $KE = \frac{1}{2}mv^2$, where $m$ is mass and $v$ is velocity. Work done (change in kinetic energy) is $\Delta KE = KE_f - KE_i$.

Step2: Calculate Initial Kinetic Energy

Given $m = 60\ kg$, $v_i = 2\ m/s$.
$KE_i = \frac{1}{2} \times 60 \times (2)^2 = 30 \times 4 = 120\ J$.

Step3: Find Final Kinetic Energy

$\Delta KE = 1800\ J$, so $KE_f = KE_i + \Delta KE = 120 + 1800 = 1920\ J$.

Step4: Solve for Final Velocity

Using $KE_f = \frac{1}{2}mv_f^2$, rearrange for $v_f$:
$v_f = \sqrt{\frac{2KE_f}{m}} = \sqrt{\frac{2 \times 1920}{60}} = \sqrt{\frac{3840}{60}} = \sqrt{64} = 8\ m/s$.

Answer:

8