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

A 60 kg bicyclist going 2 m/s increased his work output by 1,800 J. What was his final velocity?

Initial Setup

To begin solving this problem, we need to understand the principles of work and energy. The work done on an object can be calculated using the formula:

\[W = F \cdot d \cdot \cos(\theta)\]

Where: W is the work done on the object (in joules) F is the force applied to the object (in newtons) d is the displacement of the object (in meters) θ is the angle between the force and the displacement vectors

Given that the work output increased by 1,800 J, we can set up the equation:

\[W = \Delta K.E.\]

Where: ΔK.E. is the change in kinetic energy

Solving for Final Velocity

Since the cyclist initially has a velocity of 2 m/s, we can calculate the initial kinetic energy as:

\[K.E. = \frac{1}{2} m v^2\]

Substitute the values of mass (60 kg) and initial velocity (2 m/s) into the equation:

\[K.E. = \frac{1}{2} \times 60 \times (2)^2\] \[K.E. = 60 \text{ J}\]

Now, we know that the work done on the cyclist is equal to the change in kinetic energy. Therefore, we can set up the equation:

\[W = K.E.{\text{final}} K.E.{\text{initial}}\] \[1,800 = K.E._{\text{final}} 60\]

Rearranging the equation gives us the final kinetic energy:

\[K.E.{\text{final}} = 1,800 + 60\] \[K.E.{\text{final}} = 1,860 \text{ J}\]

Finally, we can solve for the final velocity using the formula for kinetic energy:

\[K.E. = \frac{1}{2} m v^2\]

Substitute the mass (60 kg) and the final kinetic energy (1,860 J) into the equation:

\[1,860 = \frac{1}{2} \times 60 \times v^2\] \[v^2 = \frac{1,860 \times 2}{60}\] \[v^2 = 62\] \[v = \sqrt{62}\] \[v \approx 7.87 \text{ m/s}\]

Conclusion

Therefore, the final velocity of the 60 kg bicyclist after increasing his work output by 1,800 J is approximately 7.87 m/s. By understanding the principles of work and energy, we were able to calculate the final velocity of the cyclist accurately.