A 4 kg weight falls on a spring that is facing up. The weight compresses the spring 0.25m. If the spring constant of the spring is 3.5N/m, how high was the weight when it was dropped?
To determine the height from which the weight was dropped, we need to use the principle of conservation of mechanical energy. When the weight is at its initial height, it has potential energy due to its height above the ground. As it falls and compresses the spring, this potential energy is converted into potential energy stored in the compressed spring.
The potential energy at the initial height is given by the formula:
Potential Energy = mass * gravitational acceleration * height
The potential energy stored in the compressed spring is given by Hooke's Law:
Potential Energy = (1/2) * spring constant * compression^2
Since the potential energy is conserved, we can equate the two expressions:
mass * gravitational acceleration * height = (1/2) * spring constant * compression^2
Now let's plug in the given values:
mass = 4 kg
gravitational acceleration = 9.8 m/s^2
compression = 0.25 m
spring constant = 3.5 N/m
Substituting these values and rearranging the equation to solve for height, we get:
height = (1/2) * spring constant * compression^2 / (mass * gravitational acceleration)
height = (1/2) * 3.5 N/m * (0.25 m)^2 / (4 kg * 9.8 m/s^2)
Simplifying the expression:
height = 0.03125 m / 39.2 m/s^2
height = 0.000798 m
Therefore, the weight was dropped from a height of approximately 0.000798 meters.