Sliding Blocks
(14 points possible)
A small cube of mass 1.0 kg slides down a circular and frictionless track of radius 0.6 m cut into a large block of mass 4.0 kg as shown in the figure below. The large block rests on a horizontal and frictionless table. The cube and the block are initially at rest, and the cube starts from the top of the path. Find the speed of the cube as it leaves the block. Take 10.0 m/s . Enter your answer in m/s.
I did the gun range on a slope problem yesterday not realizing it was a final exam question. Enough !
To find the speed of the cube as it leaves the block, we can use the principle of conservation of mechanical energy. According to this principle, the total mechanical energy of the system is conserved if there are no external forces doing work on it.
In this case, the only external force present is gravity, which does not do any work since the track is frictionless. Therefore, the mechanical energy of the system is conserved.
We can calculate the initial and final mechanical energies of the system and equate them to find the speed of the cube.
1. Initial mechanical energy (E_initial):
- Since both the cube and the block are initially at rest, their initial kinetic energy is zero.
- The only energy present is the gravitational potential energy of the cube, which can be calculated as m_cube * g * h, where m_cube is the mass of the cube, g is the acceleration due to gravity (9.8 m/s^2), and h is the height from which the cube starts.
- In this case, the height is the same as the radius of the circular track, h = 0.6 m.
- Substituting the values, the initial mechanical energy is E_initial = 1.0 kg * 9.8 m/s^2 * 0.6 m = 5.88 J.
2. Final mechanical energy (E_final):
- At the bottom of the circular track, both the cube and the block have kinetic energy.
- The kinetic energy of the cube can be calculated as (1/2) * m_cube * v^2, where v is the speed of the cube.
- The kinetic energy of the block can be calculated as (1/2) * m_block * V^2, where m_block is the mass of the block and V is the final velocity of the block (which is the same as the velocity of the cube at that point).
- Since the cube leaves the block, the final mechanical energy only consists of the kinetic energies of both the cube and the block.
- Therefore, the final mechanical energy is E_final = (1/2) * 1.0 kg * v^2 + (1/2) * 4.0 kg * v^2 = 2.5 v^2 + 2 v^2 = 4.5 v^2.
Since mechanical energy is conserved, we equate E_initial and E_final:
E_initial = E_final
5.88 J = 4.5 v^2
Solving for v^2, we get:
v^2 = 5.88 J / 4.5
v^2 = 1.307 J
Taking the square root of both sides, we get:
v = √(1.307 J) ≈ 1.142 m/s
Therefore, the speed of the cube as it leaves the block is approximately 1.142 m/s.