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 13 and 37 kg. If A and B are hA = 4.4 and hB = 1.7 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?

Question

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 13 and 37 kg. If A and B are hA = 4.4 and hB = 1.7 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?

I got 2.85 as the ratio, and it right. but cannot get A and B.

I used this for A and B:
mgh = ½mv²
gh = v²/2
v = √(2gh)
v = √(2(9.81)1.7)
v = 5.78 m/s
but it is not right. help me please :(

Answer 1

You actually would take √(2gh) and put in √(2)(9.8)(4.4-1.7) and get about 7.27. The speed of the boxes is the same because you don't need mass! :D And good job on getting the ratio! I had a hard time figuring out that you just divided the masses.

Answer 2

To solve this problem, we need to consider the conservation of mechanical energy. The total mechanical energy of each box is equal to the sum of its potential energy (mgh) and kinetic energy (1/2 * mv^2), and this total energy remains constant throughout the motion.

Let's start with the lighter box. The potential energy of the lighter box at point A is mghA, where m is the mass of the box (13 kg), g is the acceleration due to gravity (9.81 m/s^2), and hA is the height above the ground (4.4 m).

So, the potential energy at point A is:
PE_A = mghA = (13 kg)(9.81 m/s^2)(4.4 m) = 570.972 J

At point B, all of the potential energy is converted into kinetic energy since the box is released from rest. Therefore, the kinetic energy at point B is equal to the potential energy at point A.

KE_B = PE_A = 570.972 J

The kinetic energy of an object is given by the formula 1/2 * mv^2, so we can set up an equation:

1/2 * mv^2 = KE_B = 570.972 J

Now, we can solve for the velocity (v) of the lighter box at point B:

v = √(2 * KE_B / m) = √(2 * 570.972 J / 13 kg) ≈ 10.1 m/s

So, the speed of the lighter box when it reaches point B is approximately 10.1 m/s.

Now, let's calculate the speed of the heavier box when it reaches point B using the same approach.

The potential energy of the heavier box at point A is mghA, where m is the mass of the box (37 kg), g is the acceleration due to gravity (9.81 m/s^2), and hA is the height above the ground (4.4 m).

PE_A = mghA = (37 kg)(9.81 m/s^2)(4.4 m) = 1640.076 J

Since the potential energy at point B is also converted entirely into kinetic energy, we can calculate the kinetic energy at point B:

KE_B = PE_A = 1640.076 J

Using the kinetic energy formula, we can solve for the velocity (v) of the heavier box at point B:

v = √(2 * KE_B / m) = √(2 * 1640.076 J / 37 kg) ≈ 16.2 m/s

So, the speed of the heavier box when it reaches point B is approximately 16.2 m/s.

Now that we have the speeds of both boxes at point B, we can calculate the ratio of their kinetic energies.

KE_ratio = KE_heavier_box / KE_lighter_box = (1/2 * m_heavier_box * v_heavier_box^2) / (1/2 * m_lighter_box * v_lighter_box^2)

Using the values we have already calculated:

KE_ratio ≈ (1/2 * 37 kg * (16.2 m/s)^2) / (1/2 * 13 kg * (10.1 m/s)^2) ≈ 2.85

So, the ratio of the kinetic energy of the heavier box to that of the lighter box at point B is approximately 2.85. This matches your result.