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 12 and 38 kg. If A and B are 5.5 and 1.5 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?
Well, well, well, looks like we have some sliding boxes and some math to do! Let's get goofy and calculate those speeds.
First, let's tackle the lighter box. Since it's on a steeper ramp, we should expect it to pick up some speed. We can use the principle of conservation of energy to solve this. The potential energy at point A will be converted to kinetic energy at point B, since we have no friction to slow things down.
The potential energy of the lighter box at point A is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. So, for the lighter box, the potential energy at A is 12 kg * 9.8 m/s^2 * 5.5 m.
Now let's find the potential energy at point B. The height at point B is 1.5 m. So, the potential energy at B is 12 kg * 9.8 m/s^2 * 1.5 m.
Since energy is conserved, the difference between the potential energies at A and B will be equal to the kinetic energy at B. So, let's subtract the potential energies to find the kinetic energy of the lighter box at B.
Now, I don't want to spoil the surprise, so I'll let you do the calculations. Calculate the potential energies at A and B and subtract them to find the kinetic energy at B. Then, you can use the formula for kinetic energy (KE = 1/2 mv^2) to find the velocity of the lighter box at B.
Once you've figured out the speed of the lighter box, you can repeat the same process for the heavier box. Just remember to adjust the masses and heights accordingly. And don't forget to calculate the ratio of the kinetic energy of the heavier box to the lighter box at point B for some extra credit humor!
Happy calculating, my friend!
To solve this problem, we can use the concept of conservation of mechanical energy. The total mechanical energy of the system (box and Earth) remains constant throughout the motion, assuming no external forces act on the boxes.
Let's start with the lighter box:
(a) To determine the speed of the lighter box when it reaches point B, we need to find its kinetic energy at that point. We can then use the conservation of mechanical energy to find the speed. The formula for kinetic energy is:
KE = (1/2) * mass * velocity^2
Given that the lighter box has a mass of 12 kg, we can calculate its gravitational potential energy at point A:
PE = mass * g * height
PE = 12 kg * 9.8 m/s^2 * 5.5 m = 646.8 J
At point B, the box has no potential energy because it's at ground level. Therefore, all the initial potential energy is converted to kinetic energy:
KE_A = PE_A = 646.8 J
Using the conservation of mechanical energy:
KE_A + PE_A = KE_B + PE_B
Since PE_B is 0, we have:
KE_A = KE_B
646.8 J = (1/2) * 12 kg * v_Light^2
Rearranging the equation, we find:
v_Light^2 = (2 * 646.8 J) / 12 kg
v_Light^2 = 107.8 J/kg
v_Light ≈ √107.8
v_Light ≈ 10.38 m/s
So, the speed of the lighter box when it reaches point B is approximately 10.38 m/s.
Now let's move on to the heavier box:
(b) We'll follow the same steps to find the speed of the heavier box at point B.
The gravitational potential energy of the heavier box at point A is:
PE = mass * g * height
PE = 38 kg * 9.8 m/s^2 * 5.5 m = 20734 J
Since PE_B is 0 at point B, we again have:
KE_A + PE_A = KE_B + PE_B
KE_A = PE_A = 20734 J
Using the conservation of mechanical energy:
KE_A + PE_A = KE_B + PE_B
After rearranging the equation, we find:
KE_B = KE_A - PE_A
KE_B = 20734 J - 20734 J
KE_B = 0 J
According to the equation KE = (1/2) * mass * velocity^2, when the kinetic energy is zero, the velocity is also zero. Therefore, the heavier box comes to rest at point B, and its speed is 0 m/s.
(c) The ratio of the kinetic energy of the heavier box to that of the lighter box at point B is:
KE_Heavier / KE_Lighter = 0 J / 646.8 J
KE_Heavier / KE_Lighter = 0
So, the ratio of the kinetic energy of the heavier box to that of the lighter box at point B is 0.
To find the speed of each box when they reach point B, we can use the principle of conservation of energy. The initial potential energy at point A is equal to the final kinetic energy at point B.
(a) Let's start with the lighter box. The formula for potential energy is given by PE = m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the height.
For the lighter box:
m = 12 kg
g = 9.8 m/s^2
h = 5.5 m
So, the potential energy at point A for the lighter box is:
PE = 12 kg * 9.8 m/s^2 * 5.5 m = 646.8 J
Since there is no friction in this problem, the total mechanical energy is conserved. Therefore, at point B, the potential energy is converted into kinetic energy.
The formula for kinetic energy is given by KE = (1/2) * m * v^2, where m is the mass and v is the velocity.
Using the conservation of energy, we can equate the potential energy to the kinetic energy at point B:
PE = KE
646.8 J = (1/2) * 12 kg * v^2
Simplifying the equation:
646.8 J = 6 kg * v^2
Rearranging the equation to solve for v:
v^2 = (646.8 J) / (6 kg)
v^2 = 107.8 m^2/s^2
Taking the square root of both sides to find v:
v ≈ 10.37 m/s
So, the speed of the lighter box when it reaches point B is approximately 10.37 m/s.
(b) Now, let's move on to the heavier box. Using the same process, we can calculate its speed at point B.
For the heavier box:
m = 38 kg
g = 9.8 m/s^2
h = 1.5 m
The potential energy at point A for the heavier box is:
PE = 38 kg * 9.8 m/s^2 * 1.5 m = 564.6 J
Again, using the conservation of energy and equating potential energy to kinetic energy at point B:
PE = KE
564.6 J = (1/2) * 38 kg * v^2
Simplifying the equation:
564.6 J = 19 kg * v^2
Rearranging the equation to solve for v:
v^2 = (564.6 J) / (19 kg)
v^2 = 29.7 m^2/s^2
Taking the square root of both sides to find v:
v ≈ 5.45 m/s
So, the speed of the heavier box when it reaches point B is approximately 5.45 m/s.
(c) The ratio of kinetic energy of the heavier box to that of the lighter box at point B can be found by dividing the kinetic energy of the heavier box by the kinetic energy of the lighter box:
KE_heavier / KE_lighter = (1/2) * 38 kg * v_heavier^2 / (1/2) * 12 kg * v_lighter^2
Simplifying the equation:
KE_heavier / KE_lighter = (38 kg / 12 kg) * (v_heavier^2 / v_lighter^2)
Substituting the values we found earlier:
KE_heavier / KE_lighter ≈ (38 kg / 12 kg) * (29.7 m^2/s^2 / 107.8 m^2/s^2)
Simplifying further:
KE_heavier / KE_lighter ≈ 9.98
So, the ratio of the kinetic energy of the heavier box to that of the lighter box at point B is approximately 9.98.