a 65kg bicyclist (including the bicycle), initially at rest, pedals with 400w of power for 6.6s. neglect…

a 65kg bicyclist (including the bicycle), initially at rest, pedals with 400w of power for 6.6s. neglect loss. list the energy types at the initial and final time and whether work and loss (due to non - conservative forces) occur as well as the corresponding amounts of energy. initial: none = 0 j work? work = 2640 j loss? no loss = 0 j final: ke = 2640 j what was her final speed? 9.01 m/s

a 65kg bicyclist (including the bicycle), initially at rest, pedals with 400w of power for 6.6s. neglect loss. list the energy types at the initial and final time and whether work and loss (due to non - conservative forces) occur as well as the corresponding amounts of energy. initial: none = 0 j work? work = 2640 j loss? no loss = 0 j final: ke = 2640 j what was her final speed? 9.01 m/s

Answer

Explanation:

Step1: Recall kinetic - energy formula

The formula for kinetic energy is $KE=\frac{1}{2}mv^{2}$, where $m$ is the mass and $v$ is the speed. We know that the final kinetic energy $KE = 2640J$ and the mass $m = 65kg$.

Step2: Solve for the speed $v$

Starting from $KE=\frac{1}{2}mv^{2}$, we can re - arrange the formula to solve for $v$. First, multiply both sides by 2 to get $2KE=mv^{2}$. Then, divide both sides by $m$: $v^{2}=\frac{2KE}{m}$. Taking the square - root of both sides gives $v=\sqrt{\frac{2KE}{m}}$. Substitute $KE = 2640J$ and $m = 65kg$ into the formula: $v=\sqrt{\frac{2\times2640}{65}}=\sqrt{\frac{5280}{65}}\approx\sqrt{81.23}\approx9.01m/s$.

Answer:

$9.01m/s$