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?
Gravitational potential energy loss (at maximum compression) = Spring potential energy gained
(There is zero kinetic energy when the drop begins and when maximum compression is attained.)
M g H = (1/2) k *(0.027)^2
k is the spring constant, 400 N/m
H is the distance you want to know.
g = 9.8 m/s^2
M = 0.30 kg
Solve for H
To solve this problem, we can use the principle of conservation of mechanical energy. The potential energy of the block before it is dropped is converted into the potential energy stored in the compressed spring.
The potential energy of an object with mass (m) at height (h) is given by the equation:
PE = mgh
Where "g" is the acceleration due to gravity.
The potential energy stored in a compressed spring is given by the equation:
PE = (1/2)kx^2
Where "k" is the spring constant and "x" is the displacement of the spring.
In this problem, when the block is dropped, all of its potential energy is eventually converted into the potential energy stored in the compressed spring. Therefore, we can equate the two equations:
mgh = (1/2)kx^2
Rearranging the equation, we get:
h = (1/2)(k/m)x^2
Now we can substitute the given values:
k = 400 N/m
m = 0.30 kg
x = 2.7 cm = 0.027 m
Plugging these values into the equation, we can find the height (h):
h = (1/2)(400 N/m / 0.30 kg)(0.027 m)^2
h ≈ 0.0725 m
Converting this height to centimeters:
h ≈ 7.25 cm
Therefore, the block was dropped from a height of approximately 7.25 cm above the compressed spring.
To solve this problem, we need to use the principle of conservation of energy. We'll consider the potential energy of the block before it is dropped, the kinetic energy of the block just before it collides with the spring, and the potential energy stored in the compressed spring.
The potential energy of the block before it is dropped can be calculated using the formula:
Potential energy = mgh
Where:
m = mass of the block = 0.30 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height above the compressed spring (unknown)
Now, when the block collides with the spring and comes to a momentary halt, its kinetic energy is converted into potential energy stored in the compressed spring. We can calculate the kinetic energy just before the collision using the formula:
Kinetic energy = (1/2)mv^2
Where:
m = mass of the block = 0.30 kg
v = velocity of the block just before the collision (unknown)
At the moment of collision, the block is at its highest point and has momentarily come to rest. Therefore, its kinetic energy is zero.
Now, we can find the potential energy stored in the compressed spring using Hooke's Law:
Potential energy of the spring = (1/2)kx^2
Where:
k = spring constant = 400 N/m
x = compression of the spring = 2.7 cm = 0.027 m
Since the potential energy of the block is converted into potential energy of the spring, we can set up an equation:
mgh = (1/2)kx^2
Substituting the known values, we get:
0.30 kg * 9.8 m/s^2 * h = (1/2) * 400 N/m * (0.027 m)^2
Now, we can solve for h:
h = [(1/2) * 400 N/m * (0.027 m)^2] / (0.30 kg * 9.8 m/s^2)
Simplifying this expression, we find:
h ≈ 0.053 m
To express the height above the compressed spring in centimeters, we can multiply h by 100:
h ≈ 0.053 m * 100 cm/m = 5.3 cm
Therefore, the block was dropped from a height of approximately 5.3 cm above the compressed spring.