1.) A vertical spring with a spring constant of 450 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30 kg block is dropped from rest. It collides with and sticks to the spring, which is compressed 2.5 cm in bringing the block to a momentary halt. Air resistance neglible, from what height above the compressed string was the block dropped?
(Steps please?
To find the height from which the block was dropped, we can use the conservation of mechanical energy. The potential energy of the block at the initial height is equal to the sum of its gravitational potential energy and the potential energy stored in the compressed spring.
The potential energy of the block at its initial height is given by:
PE_initial = m * g * h
Where m is the mass of the block (0.30 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the initial height.
At the moment the block is compressed and comes to a halt, all of its initial potential energy has been converted into the spring potential energy, given by:
PE_spring = (1/2) * k * x^2
Where k is the spring constant (450 N/m) and x is the compression of the spring (2.5 cm or 0.025 m).
Since the block comes to a momentary halt when it collides with the spring, its kinetic energy at that moment is zero.
Since energy is conserved, we can equate the initial potential energy to the spring potential energy:
m * g * h = (1/2) * k * x^2
Rearranging the equation to solve for h:
h = (1/2) * k * x^2 / (m * g)
Now, we can substitute the given values into the equation to find the height:
h = (1/2) * (450 N/m) * (0.025 m)^2 / (0.30 kg * 9.8 m/s^2)
Calculating this expression will give us the height from which the block was dropped.