If a 75-kg skater starts his skate at 8.0m, at his lowest point (height = 0), what is his velocity?
To calculate the skater's velocity at the lowest point, we need to use the concept of conservation of energy. At the highest point (maximum height), the skater has maximum potential energy (mgh) and zero kinetic energy. At the lowest point, all the potential energy is converted into kinetic energy.
The potential energy at the highest point can be calculated using the formula: Potential Energy (PE) = mass (m) × gravitational acceleration (g) × height (h).
In this case, m = 75 kg, g = 9.8 m/s², and h = 8.0 m.
PE = 75 kg × 9.8 m/s² × 8.0 m = 5880 J.
Since all the potential energy is converted into kinetic energy at the lowest point, we can equate these two energies: Potential Energy (PE) = Kinetic Energy (KE).
So, 5880 J = 0.5 × mass (m) × velocity squared (v²).
Rearranging the equation to solve for velocity (v), we get:
v² = (5880 J) / (0.5 × 75 kg).
Dividing both sides by 0.5 × 75 kg and taking the square root will give us the velocity at the lowest point:
v = sqrt((5880 J) / (0.5 × 75 kg)).
Calculating this equation:
v = sqrt((5880 J) / (0.5 × 75 kg)) = sqrt(156.8 m²/s²).
Therefore, the velocity of the skater at the lowest point is approximately 12.5 m/s.
Is this some kind of downhill skating?
Are you supposed to ignore friction?
Is he starting the skate with zero velocity?
This is a pretty far-fetched question. I suppose they want you to use conservation of energy.
That will result in V(final) = sqrt(2gH)
where H = 8 m.
The weight won't make a difference.