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?
Fs = 5000 N/m * 0.120 m = 600 N.=Force
of the spring.
0.5m*V^2 = P.E. = Fs*d,
0.125*V^2 = 600*0.120,
0.125V^2 = 72,
V^2 = 576,
V = 24 m/s. = Vo.
h = (Vf^2-Vo^2)/2g,
h = (0-(24)^2) / -19.6 = 29.4 m.
I CAN'T GET THE RIGHT ANSWER :( A 0.230 kg block on a vertical spring with spring constant of 4.25 multiplied by 103 N/m is pushed downward, compressing the spring 0.048 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?
To solve this problem, we can use the principle of conservation of mechanical energy. The initial potential energy of the system when the block is compressed is equal to the final potential energy at the highest point of the block's trajectory.
Here are the steps to solve the problem:
Step 1: Find the potential energy stored in the spring when it is compressed.
The potential energy stored in a spring is given by the formula: PE = (1/2) k x^2
where PE is the potential energy, k is the spring constant, and x is the amount of compression or elongation of the spring.
Given:
k = 5.00 × 10^3 N/m (spring constant)
x = 0.120 m (compression)
Using the formula, we can calculate the potential energy stored in the spring:
PE = (1/2) × 5.00 × 10^3 N/m × (0.120 m)^2
PE = 36 J
So, the potential energy stored in the spring when it is compressed is 36 J.
Step 2: Find the maximum height the block reaches above the point of release.
At the highest point of the block's trajectory, all of its potential energy is converted to gravitational potential energy.
Gravitational potential energy is given by the formula: PE = mgh
where m is the mass, g is the acceleration due to gravity, and h is the height.
Given:
m = 0.250 kg (mass)
g = 9.8 m/s^2 (acceleration due to gravity)
PE = 36 J (potential energy)
Using the formula, we can calculate the maximum height:
36 J = 0.250 kg × 9.8 m/s^2 × h
h = 36 J / (0.250 kg × 9.8 m/s^2)
h = 14.6 m
Therefore, the block rises 14.6 meters above the point of release.
Final Answer: The block rises 14.6 meters above the point of release.
To determine how high the block rises above the point of release, we can use the principle of conservation of mechanical energy. The total mechanical energy before release is equal to the total mechanical energy after release.
First, let's find the total mechanical energy before release (at the compressed point). The total mechanical energy consists of potential energy (PE) and kinetic energy (KE).
PE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height.
KE = 0.5mv^2, where m is the mass of the block, and v is the velocity.
At the compressed point, the block is at its maximum potential energy because it is at its maximum height. Therefore, KE is equal to zero.
PE = 0.250 kg * 9.8 m/s^2 * 0.120 m = 0.294 J
The total mechanical energy before release is 0.294 J.
Next, let's find the total mechanical energy after release (at the highest point the block reaches). At the highest point, the block momentarily stops, so the velocity is zero. Therefore, KE is zero.
PE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height.
Since the velocity is zero at the highest point, all the initial potential energy is converted into gravitational potential energy as the block rises.
PE = 0.294 J
0.294 J = 0.250 kg * 9.8 m/s^2 * h
Solving for h, we find h = 0.294 J / (0.250 kg * 9.8 m/s^2) = 0.120 m
Therefore, the block rises 0.120 m above the point of release.