A vertical spring with a spring constant of 400 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 by 2.7 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in cm) above the compressed spring was the block dropped?
To find the height from which the block was dropped, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system (block + spring) is conserved throughout the motion.
The total mechanical energy is the sum of the potential energy (PE) and the kinetic energy (KE).
Initially, when the block is at rest, it only has potential energy at height h:
PE_initial = m * g * h
Where:
m = mass of the block (0.30 kg)
g = acceleration due to gravity (9.8 m/s²)
When the block lands on the spring, the potential energy is converted into the elastic potential energy stored in the compressed spring.
The elastic potential energy stored in a spring is given by the equation:
PE_spring = (1/2) * k * x²
Where:
k = spring constant (400 N/m)
x = compression of the spring (2.7 cm = 0.027 m)
Finally, when the block comes to a momentary halt, all its energy is stored in the compressed spring. At this point, the block is at its maximum compression, so its potential energy is zero.
Equating the initial potential energy to the spring potential energy, we have:
PE_initial = PE_spring
m * g * h = (1/2) * k * x²
Plugging in the given values:
(0.30 kg) * (9.8 m/s²) * h = (1/2) * (400 N/m) * (0.027 m)²
Now, solve for h:
h = [(1/2) * (400 N/m) * (0.027 m)²] / [(0.30 kg) * (9.8 m/s²)]
Calculate the value:
h ≈ 3.29 meters or 329 cm
Therefore, the block was dropped from a height of approximately 329 cm above the compressed spring.