A spring with k= 124 N/m is compressed a distance x= 15 cm. This compressed posi

Question

A spring with k= 124 N/m is compressed a distance x= 15 cm. This compressed position is defined to be at a height of zero. A mass with m= 30 grams is placed on top of the compressed spring (still at zero height). When the spring is released, it pushes upward on the mass, launching it to a height h above it's initial position. How high does the mass go? (Hint: The total kinetic + spring + gravitational energy is conserved. And the kinetic energy is zero just before launch and at the top of the trajectory, so that only the spring and gravitational energies need to be considered.) A spring with k= 124 N/m is compressed a distance x= 15 cm. This compressed position is defined to be at a height of zero. A mass with m= 30 grams is placed on top of the compressed spring (still at zero height). When the spring is released, it pushes upward on the mass, launching it to a height h above it's initial position. How high does the mass go? (Hint: The total kinetic + spring + gravitational energy is conserved. And the kinetic energy is zero just before launch and at the top of the trajectory, so that only the spring and gravitational energies need to be considered.) A spring with k= 124 N/m is compressed a distance x= 15 cm. This compressed position is defined to be at a height of zero. A mass with m= 30 grams is placed on top of the compressed spring (still at zero height). When the spring is released, it pushes upward on the mass, launching it to a height h above it's initial position. How high does the mass go? (Hint: The total kinetic + spring + gravitational energy is conserved. And the kinetic energy is zero just before launch and at the top of the trajectory, so that only the spring and gravitational energies need to be considered.)

Explanation / Answer

PE of the spring = PE of the block

0.5kx^2 = mg(h + x)

0.5*124*0.15^2 = 0.03*9.8*(h + 0.15)

h = 4.594 m

so the mass goes to a height of 4.594 m

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