A small cube of mass m1= 1.0 kg slides down a circular and frictionless track of radius R= 0.5 m cut into a large block of mass m2= 4.0 kg. The large block rests on a horizontal and frictionless table. The cube and the block are initially at rest, and the cube m1 starts from the top of the path. Find the speed of the cube v1 as it leaves the block. Take g= 10.0 m/s2. Give your answer in m/s.
I just figured it out :)
To find the speed of the cube as it leaves the block, we can use the principle of conservation of mechanical energy.
The system consists of the cube of mass m1 and the block of mass m2. Initially, both objects are at rest, so their initial kinetic energy is zero.
The cube slides down the circular track, gaining gravitational potential energy as it descends. When it leaves the block, it will have converted all of its potential energy into kinetic energy.
The formula for gravitational potential energy is given by U = m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the height.
Since the cube starts from the top of the path, its initial height is equal to the radius of the circular track, R.
The gravitational potential energy at the top, U1, is given by:
U1 = m1 * g * R
When the cube leaves the block, it will have converted all of its potential energy into kinetic energy.
The formula for kinetic energy is given by K = (1/2) * m * v^2, where m is the mass and v is the velocity.
Equating the initial potential energy to the final kinetic energy, we have:
U1 = K1
m1 * g * R = (1/2) * m1 * v1^2
Simplifying and solving for v1, we get:
v1^2 = 2 * g * R
Substituting the given values, we have:
v1^2 = 2 * 10 * 0.5
v1^2 = 10
Taking the square root of both sides, we find:
v1 ≈ 3.16 m/s
Therefore, the speed of the cube as it leaves the block is approximately 3.16 m/s.