The drawing shows two boxes resting on frictionless ramps. One box is relatively light and sits on a steep ramp. The other box is heavier and rests on a ramp that is less steep. The boxes are released from rest at A and allowed to slide down the ramps. The two boxes have masses of 9 and 48 kg. If A and B are hA = 4.8 and hB = 1.2 m, respectively, above the ground, determine the speed of (a) the lighter box and (b) the heavier box when each reaches B. (c) What is the ratio of the kinetic energy of the heavier box to that of the lighter box at B?
a. Box 1
V^2 = Vo^2+2g*h = 0 + 19.6*(4.8-1.2) =
70.56
V = 8.4 m/s.
b. Box 2.
V^2 = Vo^2+2g*h = 0 + 19.6*(4.8-1.2) =
70.56
V = 8.4 m/s.
c. KE2/KE1 = 0.5m2*V^2/0.5m1*V^2 =
24*70.56/4.5*70.56 = 5.33 = m2/m1.
To solve this problem, we can use the principles of conservation of mechanical energy. The concept states that the total mechanical energy (sum of potential energy and kinetic energy) of an object is constant if no external forces are acting on it.
Given:
Mass of the lighter box (m1) = 9 kg
Mass of the heavier box (m2) = 48 kg
Height at A for the lighter box (hA1) = 4.8 m
Height at A for the heavier box (hA2) = 1.2 m
(a) To find the speed of the lighter box at point B, we can equate the potential energy at point A to the kinetic energy at point B.
Potential energy at A (PEA1) = m1 * g * hA1
Kinetic energy at B (KEB1) = (1/2) * m1 * v1^2
Setting these equal and solving for v1:
m1 * g * hA1 = (1/2) * m1 * v1^2
v1 = √(2 * g * hA1)
(b) Similarly, to find the speed of the heavier box at point B, we use the same equation but with the mass of the heavier box.
Potential energy at A (PEA2) = m2 * g * hA2
Kinetic energy at B (KEB2) = (1/2) * m2 * v2^2
Setting these equal and solving for v2:
m2 * g * hA2 = (1/2) * m2 * v2^2
v2 = √(2 * g * hA2)
(c) The ratio of kinetic energy of the heavier box to that of the lighter box at B can be calculated as:
KEB2 / KEB1 = (1/2) * m2 * v2^2 / (1/2) * m1 * v1^2
= m2 * v2^2 / m1 * v1^2
Now, substitute the values of m1, m2, v1, and v2 to find the ratio.
Please note that g is the acceleration due to gravity (approximately 9.8 m/s^2).
To find the speed of the boxes when they reach point B, we can use the principle of conservation of mechanical energy. The total mechanical energy of each box at point A is equal to its potential energy, which is given by the product of its mass, the acceleration due to gravity, and the height of the ramp.
(a) For the lighter box with a mass of 9 kg, the potential energy at point A is:
E(A) = m * g * hA
= 9 kg * 9.8 m/s^2 * 4.8 m
= 423.36 J
Using the conservation of mechanical energy, we can equate the initial potential energy at point A to the final mechanical energy at point B, which is the sum of the kinetic energy and potential energy at point B.
E(A) = E(B)
At point B, the potential energy is given by:
E(B) = m * g * hB
Substituting the values, we get:
423.36 J = 9 kg * 9.8 m/s^2 * 1.2 m + 1/2 * 9 kg * v^2
Rearranging the equation to solve for v, we get:
v^2 = (2 * (423.36 J - 9 kg * 9.8 m/s^2 * 1.2 m)) / 9 kg
v^2 = (2 * (423.36 J - 105.84 J)) / 9 kg
v^2 = (2 * 317.52 J) / 9 kg
v^2 = 70.56 J / 9 kg
v^2 = 7.84 m^2/s^2
Taking the square root of both sides, we find:
v = √7.84 m/s
v ≈ 2.8 m/s
Therefore, the speed of the lighter box when it reaches point B is approximately 2.8 m/s.
(b) Similarly, we can calculate the speed of the heavier box with a mass of 48 kg using the same approach.
E(A) = m * g * hA
= 48 kg * 9.8 m/s^2 * 4.8 m
= 2252.16 J
Using the conservation of mechanical energy:
2252.16 J = 48 kg * 9.8 m/s^2 * 1.2 m + 1/2 * 48 kg * v^2
Rearranging and solving for v, we get:
v^2 = (2 * (2252.16 J - 48 kg * 9.8 m/s^2 * 1.2 m)) / 48 kg
v^2 = (2 * (2252.16 J - 564.48 J)) / 48 kg
v^2 = (2 * 1687.68 J) / 48 kg
v^2 = 70.32 J / 48 kg
v^2 = 1.46 m^2/s^2
Finding the square root of both sides, we find:
v = √1.46 m/s
v ≈ 1.21 m/s
Therefore, the speed of the heavier box when it reaches point B is approximately 1.21 m/s.
(c) The ratio of the kinetic energy of the heavier box to that of the lighter box at point B can be calculated by dividing their respective kinetic energies.
The kinetic energy of the lighter box at point B is given by:
K(lighter) = 1/2 * m * v(lighter)^2
= 1/2 * 9 kg * (2.8 m/s)^2
= 35.28 J
The kinetic energy of the heavier box at point B is given by:
K(heavier) = 1/2 * m * v(heavier)^2
= 1/2 * 48 kg * (1.21 m/s)^2
= 35.47 J
Therefore, the ratio of the kinetic energy of the heavier box to that of the lighter box at point B is approximately 35.47 J / 35.28 J, which simplifies to approximately 1.006.