A 1kg mass is suspended from a vertical spring with a spring constant of 100 N/m and the equilibrium position is noted. The spring is then pushed upward (compressed) a distance x=10 cm before the mass is released from rest. How fast will the block be moving when it passes through the equilibrium position?
Potential energy stored in the spring during compression will be (1/2)k(0.1)^2
= 0.5*100*10^-2 = 0.5 J. There will be additional stored gravitational energy of 1 kg*9.8 m/s^2*0.1 m = 1.0 J
When the block passes through the equilibrium position, 1.5 will be converted to kinetic energy, (1/2) m V^2
Solve for V
To solve this problem, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system (spring-mass system) is conserved throughout the motion.
The total mechanical energy is the sum of the potential energy and the kinetic energy:
Total mechanical energy = Potential energy + Kinetic energy
1. Potential Energy: The potential energy of a spring can be given by the formula:
Potential energy of the spring = (1/2) * k * x^2
Where k is the spring constant and x is the displacement from the equilibrium position. In this case, the displacement is given as x = 10 cm = 0.1 m.
Potential energy of the spring = (1/2) * 100 N/m * (0.1 m)^2 = 0.5 J
2. Kinetic Energy: At the equilibrium position, the block comes to rest momentarily. Hence, the kinetic energy at the equilibrium position is zero.
Now, at the moment the block passes through the equilibrium position, all the potential energy will be converted into kinetic energy.
Total mechanical energy = Potential energy + Kinetic energy
Since the potential energy is 0.5 J at the equilibrium position, the kinetic energy will also be 0.5 J at that moment.
Now, we can use the formula for kinetic energy to find the velocity:
Kinetic energy = (1/2) * mass * velocity^2
Solving for velocity:
0.5 J = (1/2) * 1 kg * velocity^2
velocity^2 = 1 J/kg
velocity = sqrt(1 J/kg)
velocity ≈ 1 m/s
Therefore, the block will be moving at approximately 1 m/s when it passes through the equilibrium position.
To determine the speed of the block when it passes through the equilibrium position, we can make use of the principle of conservation of mechanical energy.
1. First, we need to find the potential energy stored in the compressed spring.
The potential energy stored in a spring can be calculated using the formula:
Potential energy = 0.5 * k * x^2
where k is the spring constant and x is the displacement of the spring from its equilibrium position.
Plugging in the values:
Potential energy = 0.5 * 100 N/m * (10 cm / 100 cm/m)^2 = 0.5 * 100 N/m * (0.1 m)^2 = 0.5 J
2. Next, we need to find the maximum kinetic energy that the block will have when it reaches the equilibrium position. At this point, all the potential energy is converted into kinetic energy.
The equation for kinetic energy is given by:
Kinetic energy = 0.5 * m * v^2
where m is the mass of the block and v is its velocity.
Since the potential energy is converted entirely into kinetic energy, we can equate the two:
0.5 J = 0.5 * 1 kg * v^2
Solving for v^2, we get:
v^2 = 0.5 J / (0.5 kg) = 1 J/kg
Taking the square root of both sides, we find:
v = sqrt(1 J/kg) = 1 m/s
So, the block will be moving at a speed of 1 m/s when it passes through the equilibrium position.