A spring with a spring constant of 224.3 N/m is compressed by 0.253 m. Then a steel ball bearing of mass 0.0433 kg is put against the end of the spring, and the spring is released. What is the speed of the ball bearing right after it loses contact with the spring? (The ball bearing will come off the spring exactly as the spring returns to its equilibrium position. Assume that the mass of the spring can be neglected.)
1/2 k x^2=1/2 m v^2 solve for v
Well, this sounds like a "spring-tacular" situation! Now, to find the speed of the ball bearing, we can use the principles of conservation of energy.
When the spring is compressed, it stores potential energy given by the equation:
Potential energy = 0.5 * k * x^2
where k is the spring constant and x is the displacement or compression of the spring.
So, the potential energy stored in the spring is:
Potential energy = 0.5 * 224.3 N/m * (0.253 m)^2 = 7.215 Joules (rounded to 3 decimal places).
This potential energy is then converted into kinetic energy as the ball bearing is released. We can equate the potential energy to the kinetic energy using the equation:
Potential energy = Kinetic energy
0.5 * m * v^2 = 7.215 Joules
Here, m is the mass of the ball bearing and v is its final speed.
Plugging in the values we have:
0.5 * 0.0433 kg * v^2 = 7.215 Joules
Now, let's solve for v:
v^2 = (2 * 7.215 Joules) / 0.0433 kg
v^2 = 332.99 m^2/s^2
v ≈ 18.247 m/s (rounded to 3 decimal places)
So, after the ball bearing loses contact with the spring, it will be "springing" away at a speed of approximately 18.247 m/s. Keep bouncing on, my friend!
To find the speed of the ball bearing right after it loses contact with the spring, we can use the principle of energy conservation. The potential energy stored in the spring when compressed will be converted into the kinetic energy of the ball bearing when it loses contact with the spring.
1. Determine the potential energy stored in the spring:
Potential energy = (1/2) * k * x^2
Where k is the spring constant and x is the compression distance.
Plugging in the values, potential energy = (1/2) * 224.3 N/m * (0.253 m)^2
2. Since the potential energy will be converted into kinetic energy, we can equate potential energy to kinetic energy:
Potential energy = Kinetic energy
(1/2) * k * x^2 = (1/2) * m * v^2
Where m is the mass of the ball bearing and v is its velocity.
3. Rearrange the equation to solve for velocity:
v^2 = (k * x^2) / m
v = √((k * x^2) / m)
Plugging in the values, v = √((224.3 N/m * (0.253 m)^2) / 0.0433 kg)
4. Calculate the value of v:
v = √(14.358 N * m^2/kg) ≈ √(14.36) m/s ≈ 3.787 m/s
Therefore, the speed of the ball bearing right after it loses contact with the spring is approximately 3.787 m/s.
To find the speed of the ball bearing right after it loses contact with the spring, we can use the principle of conservation of mechanical energy. The energy stored in the spring when it is compressed will be converted into the kinetic energy of the ball bearing when it is released.
First, let's find the potential energy stored in the compressed spring using Hooke's Law:
Potential Energy (PE) = (1/2) * k * x^2
where k is the spring constant and x is the distance the spring is compressed.
PE = (1/2) * 224.3 N/m * (0.253 m)^2
= 7.176 J
Next, let's equate the potential energy stored in the spring to the kinetic energy of the ball bearing when it loses contact with the spring:
Potential Energy (PE) = Kinetic Energy (KE)
KE = 7.176 J
The formula for kinetic energy is given by:
KE = (1/2) * m * v^2
where m is the mass of the ball bearing and v is its velocity.
Now let's solve for v:
v^2 = (2 * KE) / m
v^2 = (2 * 7.176 J) / 0.0433 kg
v^2 = 331.71
v ≈ 18.2 m/s
So, the speed of the ball bearing right after it loses contact with the spring is approximately 18.2 m/s.