Answer:
To solve this problem, we can use the conservation of energy principle. The potential energy at A is equal to the kinetic energy at B, C, and D. Since the chute is smooth, there is no friction and no energy is lost due to work done against frictional forces.
At point A, the package has only potential energy, which is given by mgh, where m is the mass of the package, g is the acceleration due to gravity, and h is the height of point A above the horizontal plane. The potential energy at A is converted to kinetic energy at point B, so we can equate these two energies to find the velocity of the package at B.
mgh = (1/2)mvB^2
where vB is the velocity of the package at point B.
Solving for vB, we get:
vB = sqrt(2gh)
Substituting the given values, we get:
vB = sqrt(2*32.2 ft/s^2 * 10 ft) = 20 ft/s
At point C, the potential energy is still the same, but the kinetic energy is different because the velocity of the package has changed. We can equate the potential energy at A to the kinetic energy at point C to find the velocity of the package at C.
mgh = (1/2)mvC^2
where vC is the velocity of the package at point C.
Solving for vC, we get:
vC = sqrt(2gh/3)
Substituting the given values, we get:
vC = sqrt(2*32.2 ft/s^2 * 10 ft/3) = 19 ft/s
At point D, the potential energy is still the same, but the kinetic energy is different because the velocity of the package has changed again. We can equate the potential energy at A to the kinetic energy at point D to find the velocity of the package at D.
mgh = (1/2)mvD^2
where vD is the velocity of the package at point D.
Solving for vD, we get:
vD = sqrt(2gh/2)
Substituting the given values, we get:
vD = sqrt(2*32.2 ft/s^2 * 10 ft/2) = 22.6 ft/s
To find the normal force at point C, we need to resolve the weight of the package into its components along the normal