A spring with a spring constant of 223.9 N/m is compressed by 0.220 m. Then a steel ball bearing of mass 0.0372 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.)
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Step 1: Calculate (2/m)
=> 2*0.0372 kg = 53.76344086 kg
Step 2: Calculate [k*({x-initial}^2)]/2
=> (223.9 N/m)*(0.220^2)= 10.41838
=> 10.41838/2 = 5.41838
Step 3: Calculate (mgh)
=> 0.0372 kg*9.81 m/s^2*0 = 0
Step 4: Add Step 2 and Step 3
=> 5.41838 + 0 = 5.41838
Step 5: Multiply Step 1 and Step 4
=> 53.76344086 kg*5.41838=291.3107527
Step 6: Square root Step 5 to get velocity
=> sqroot(291.3107527) = 17.067828 m/s
Step 7: Consider significant figures
=> 17.07 m/s
Had to break it down. The actual formula is difficult to type without confusing you, but here it is:
V= sqroot[(2/m)*{((1/2)*k*(x-initial^2))-(mgh)}
To find the speed of the ball bearing right after it loses contact with the spring, we can use the concept of conservation of mechanical energy. The initial potential energy stored in the compressed spring is converted into kinetic energy of the ball bearing.
The potential energy stored in a spring is given by the formula:
Potential Energy = (1/2) * k * x^2
where k is the spring constant and x is the compression or extension of the spring.
In this case, the spring constant is 223.9 N/m and the compression of the spring is 0.220 m. Therefore, the initial potential energy stored in the spring is:
Potential Energy = (1/2) * 223.9 N/m * (0.220 m)^2
Next, we can equate the potential energy to the kinetic energy of the ball bearing. The kinetic energy of an object is given by the formula:
Kinetic Energy = (1/2) * m * v^2
where m is the mass of the object and v is its velocity.
In this case, the mass of the steel ball bearing is 0.0372 kg. Therefore, the initial potential energy of the spring is equal to the initial kinetic energy of the ball bearing:
(1/2) * k * x^2 = (1/2) * m * v^2
We can rearrange this equation to solve for v:
v = sqrt((k * x^2) / m)
Substituting the given values, we can calculate the speed of the ball bearing:
v = sqrt((223.9 N/m * (0.220 m)^2) / 0.0372 kg)
Simplifying this expression, we get:
v = sqrt(580.976 m^2/s^2) = 24.12 m/s
Therefore, the speed of the ball bearing right after it loses contact with the spring is approximately 24.12 m/s.