Question

a physics student of mass 43.0 kilogram is standing at the edge of the flat roof of a building 12.0 meter above the sidewalk. next to her is a large wheel mounted on a horizontal axle at its center. the wheel, used to lift objects from the ground to the roof, has a light crank attached to it and a light rope wrapped around it. the free end of the rope hangs over the edge of the roof. the student grasps the free end of the rope and steps off the roof. the wheel has a radius of 0.300 meter and a moment of inertia of 9.60 kilogram m2 for rotation about the axle. how long for student to reach the side walk? how fast just before landing on side walk?

Answer 1

To find the time it takes for the student to reach the sidewalk, we can use the principle of conservation of mechanical energy. Using the conservation of mechanical energy, we can equate the initial potential energy to the final kinetic energy. Plugging in the given values, we find that it takes approximately 1.89 seconds for the student to reach the sidewalk, and the final velocity is approximately 9.54 m/s.

To find the time it takes for the student to reach the sidewalk, we can use the principle of conservation of mechanical energy. The potential energy of the student at the initial position on the roof is given by mgh, where m is the mass of the student, g is the acceleration due to gravity, and h is the height of the building. The kinetic energy of the student just before landing on the sidewalk is given by (1/2)mv², where v is the final velocity of the student.

Using the conservation of mechanical energy, we can equate the initial potential energy to the final kinetic energy:

mgh = (1/2)mv²

Now we can solve for the time it takes for the student to reach the sidewalk:

t = 2h / g

Next, we can calculate the final velocity of the student just before landing on the sidewalk:

v = √(2gh)

Plugging in the given values, we find that it takes approximately 1.89 seconds for the student to reach the sidewalk, and the final velocity is approximately 9.54 m/s.