Answer:
9.9 m/s
Step-by-step explanation:
To solve this problem, we can use conservation of energy. At point A, the box has potential energy due to its position on the hill. As it rolls down the hill, this potential energy is converted into kinetic energy, which the box has at point B.
The potential energy of the box at point A is given by mgh, where m is the mass of the box, g is the gravitational acceleration, and h is the height of the box above some reference level. The height difference between points A and B is 5 meters, so the potential energy at point A is mgh = 10.0 kg * 9.8 m/s^2 * 15 m = 1470 J.
At point B, all of the potential energy at point A has been converted into kinetic energy. The kinetic energy of the box is given by (1/2)mv^2, where m is the mass of the box and v is its velocity. Equating the potential and kinetic energies, we have:
mgh = (1/2)mv^2
Solving for v, we get:
v = sqrt(2gh)
Plugging in the values, we get:
v = sqrt(2 * 9.8 m/s^2 * 5 m) = 9.9 m/s
Therefore, the velocity of the box when it reaches point B is 9.9 m/s.
ALLEN