Question

Let’s consider the following situation: Imagine a 37 kg mass falling from a height h and landing on a spring, compressing it, and then being launched back upwards. If the spring has a spring constant of 179 N/m and is observed to be maximally compressed by 5 meters, determine the following: [Note: You may consider the top of the fully compressed spring to be located at y = 0 m] i The total energy contained in the system. ii The height above the ground from which the mass fell. iii The velocity with which the mass impacts the spring. iv The velocity of the mass when it’s launched back upwards. v What is the value of the mass’ maximum kinetic energy and at what point or points in its motion is it achieved? vi What is the value of the mass’ minimum kinetic energy and at what point or points in its motion is it achieved?

Answer 1

The total energy contained in the system

E = 1/2 * k * x²

E = 1/2 * 179 * 5²

E = 2238 J

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The height above the ground from which the mass fell

use conservation of energy

mgh = 2238 J ( assuming there is no energy lost)

also, y = 0 is at ground

so,

h = 2238 / 37 * 9.8

h = 6.17 m

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The velocity with which the mass impacts the spring

1/2 * m * v² = 2238

so,

v = sqrt ( 2 * 2238 / m)

v = 11 m/s

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The velocity of the mass when it’s launched back upwards

it will be launched when spring is at its natural ( fully stretched position)

It should be same as velocity of impact as we are talking about a system here and there is no external force

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What is the value of the mass’ maximum kinetic energy and at what point or points in its motion is it achieved

maximum K.E =  2238 J

It is obtained at equilibrium ( when spring is fully expanded)

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minimum K.E is the top of trajectory and as the maximum compression.