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A 5.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The spring constant of the spring is 1.00 N/cm. The mass is pulled downward 2.00 cm and released. What is the speed of the mass when it is 1.00 cm above the point from which it was released?

A 5.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The spring constant of the spring is 1.00 N/cm. The mass is pulled downward 2.00 cm and released. What is the speed of the mass when it is 1.00 cm above the point from which it was released?

0.1084 m/s

0.0775 m/s

0.1239 m/s

0.0930 m/s

The mass will not reach the height specified.

### 1 Answer

- NCSLv 72 years agoFavorite Answer
k = 100 N/m

At the equilibrium position the spring is extended by

x = mg / k = 5.00kg * 9.8m/s² / 100N/m = 0.49 m

After pulling down 0.0200 m,

SPE = ½ * 100N/m * (0.51m)² = 13.005 J

and after rising 1 cm,

SPE = ½ * 100N/m * (0.50m)² = 12.5 J

so it has lost 0.505 J of spring energy.

After rising that 1 cm,

GPE = mgh = 5.00kg * 9.8m/s² * 0.0100m = 0.49 J

and so

KE = 0.505J - 0.49J = 0.015 J = ½ * 5.00kg * v²

which solves to

v = 0.0775 m/s ◄ corrected typo

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