Hung vertically, a mass less spring extends by 3.00 cm a mass of 739. G is attached to its lower end. The same mass and spring are then placed apart on a table. The spring is fixed in place and then the mass is given a velocity of 0.900 m/s towards the spring.
a) What is the maximum compression of the spring when the mass runs into it?
b) After compressing the spring the mass will rebound. What is its velocity just as leave contact with the spring?
To answer these questions, we need to analyze the conservation of mechanical energy in both situations.
a) To find the maximum compression of the spring when the mass runs into it, we need to equate the potential energy of the spring with the initial kinetic energy of the mass.
The potential energy of a spring can be calculated using the formula:
Potential energy = (1/2) * k * x^2
Where:
k is the spring constant (stiffness of the spring)
x is the displacement of the spring from its equilibrium position.
In this case, the spring is initially extended by 3.00 cm, which is equal to 0.03 m, and the mass of the object is 739 g, which is equal to 0.739 kg.
First, we need to find the spring constant, k. The force exerted by a spring is given by Hooke's law, which states that F = -k * x, where F is the force, k is the spring constant, and x is the displacement.
In this case, F = m * g, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2).
So, we can write:
m * g = k * x
Plugging in the values, we get:
(0.739 kg) * (9.8 m/s^2) = k * (0.03 m)
Now we can solve for k:
k = (0.739 kg * 9.8 m/s^2) / (0.03 m)
k = 241.63 N/m
Now that we have the spring constant, we can calculate the potential energy of the spring:
Potential energy = (1/2) * k * x^2
Potential energy = (1/2) * 241.63 N/m * (0.03 m)^2
Potential energy = 0.10922 J
Since the initial kinetic energy of the mass is zero (as it starts from rest), the entire potential energy of 0.10922 J will be converted into the maximum compression of the spring. Therefore, the maximum compression of the spring is 0.10922 J.
b) To find the velocity of the mass just as it leaves contact with the spring, we can again use the conservation of mechanical energy.
The total mechanical energy of the system is conserved, so the sum of kinetic energy and potential energy before compression should equal the sum of kinetic energy and potential energy after compression.
Initially, the mass has a kinetic energy given by:
Initial kinetic energy = (1/2) * m * v^2
Where:
m is the mass of the object (0.739 kg)
v is the initial velocity of the object (0.900 m/s)
The potential energy of the spring is given by:
Potential energy = (1/2) * k * x^2
We already calculated the spring constant, k, and the maximum compression of the spring, x, is also known (0.10922 J).
Now, we can equate the initial kinetic energy with the sum of potential energy and final kinetic energy:
(1/2) * m * v^2 = (1/2) * k * x^2 + (1/2) * m * v_final^2
Since we are interested in finding the final velocity right as the mass leaves contact with the spring, we know that the potential energy of the spring will be zero at that point. Therefore, the equation simplifies to:
(1/2) * m * v^2 = (1/2) * m * v_final^2
To solve for v_final, we rearrange the equation:
v_final^2 = v^2
Taking the square root of both sides, we get:
v_final = v
Therefore, the velocity of the mass just as it leaves contact with the spring is 0.900 m/s.