# Physics

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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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• Physics -

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• Physics -

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)}

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