A 5 meter, 100 N/m spring is at a 30 degree angle to the horizontal. If a 10 kg weight is placed on the spring when it is compressed 2 meters, how high will it travel?
The initial stored potential energy in the spring,
(1/2) k X^2,
is initial converted to 3/4 (cos 30)^2 kinetic energy of horizontal motion and 1/4 (sin 30)^2 kinetic energy of vertical motion. That means (1/8) k X^2 is available to be converted to potential energy. That tells you how far it will rise above the initially compressed position. The length of the spring doesn't matter here.
How do you solve for the weight's maximum height above the ground?
To calculate how high the weight will travel after being released from the compressed spring, we can use energy conservation principles. The potential energy stored in the compressed spring will be converted into kinetic energy of the weight as it moves upwards. Once the weight reaches its highest point, all of its energy will be in the form of gravitational potential energy.
First, let's calculate the potential energy stored in the spring when it is compressed by 2 meters. The formula for potential energy stored in a spring is given by:
Potential Energy (PE) = (1/2) * k * x^2
where k is the spring constant (100 N/m) and x is the displacement from the equilibrium position (2 meters). Plugging in the values:
PE = (1/2) * (100 N/m) * (2 m)^2
= 200 J
Now, when the weight is released, all of this potential energy will be converted into gravitational potential energy at its highest point. The formula for gravitational potential energy is:
PE_gravitational = m * g * h
where m is the mass (10 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum height attained. Rearranging the equation, we can solve for h:
h = PE_gravitational / (m * g)
Plugging in the values:
h = 200 J / (10 kg * 9.8 m/s^2)
≈ 2.04 m
Therefore, the weight will travel approximately 2.04 meters high.