A 0.250 kg block on a vertical spring with a spring constant of 5.00 103 N/m is pushed downward, compressing the spring 0.120 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?
m
Equate the stored spring potential energy to the potential energy change at the highest elevation.
(1/2)k*X^2 = M g H
Solve for H
h = 10.20408163 m
h = 10 m
Why did the block go up the spring? Because it wanted to spring into action! Well, let's calculate how high it rises above the point of release.
First, let's find the potential energy stored in the compressed spring. The potential energy stored in a spring is given by the equation: 1/2 * k * x^2, where k is the spring constant and x is the displacement of the spring.
Plugging in the values, we have:
Potential energy = 1/2 * 5000 N/m * (0.120 m)^2
Now, let's equate this potential energy to the gravitational potential energy when the block reached its maximum height. The gravitational potential energy is given by the equation: m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the height.
Setting the potential energy equal to the gravitational potential energy, we have:
1/2 * 5000 N/m * (0.120 m)^2 = 0.250 kg * 9.8 m/s^2 * h
Now, let's solve for h:
h = (1/2 * 5000 N/m * (0.120 m)^2) / (0.250 kg * 9.8 m/s^2)
Calculating this, we find that the block rises approximately 0.200 meters above the point of release. So, the block sure went "springing" up high!
To determine how high the block rises above the point of release, we can use the principle of conservation of mechanical energy.
The initial potential energy stored in the spring when it is compressed is given by:
PE_initial = (1/2) * k * x^2
Where:
k = spring constant = 5.00 * 10^3 N/m
x = compression of the spring = 0.120 m
PE_initial = (1/2) * (5.00 * 10^3 N/m) * (0.120 m)^2
= 36 J
According to the conservation of mechanical energy, this potential energy will be converted into kinetic energy as the block leaves the spring and rises vertically.
When the block reaches its highest point, it comes to rest momentarily, so all the initial potential energy is converted to gravitational potential energy.
The gravitational potential energy at the highest point is given by:
PE_gravitational = m * g * h
Where:
m = mass of the block = 0.250 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height above the point of release
Setting the two potential energies equal:
PE_initial = PE_gravitational
36 J = (0.250 kg) * (9.8 m/s^2) * h
Solving for h:
h = 36 J / [(0.250 kg) * (9.8 m/s^2)]
h = 14.693 m
Therefore, the block rises approximately 14.693 meters above the point of release.
To find the height the block rises above the point of release, we need to use the law of conservation of mechanical energy.
First, let's find the potential energy stored in the compressed spring:
Potential energy (PE) = 1/2 * k * x^2
where k is the spring constant and x is the displacement of the spring (0.120 m).
PE = 0.5 * 5.00 * 10^3 N/m * (0.120 m)^2
PE = 36 J
Since the block leaves the spring and travels upward, all the potential energy from the compressed spring will be converted into gravitational potential energy.
Gravitational potential energy (PE) = m * g * h
where m is the mass of the block (0.250 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the point of release.
36 J = 0.250 kg * 9.8 m/s^2 * h
36 J = 2.45 kg m^2/s^2 * h
To find h, we can rearrange the equation:
h = 36 J / (0.250 kg * 9.8 m/s^2)
h ≈ 14.69 m
Therefore, the block will rise approximately 14.69 meters above the point of release.