A spring with spring constant k=620 N / m is placed in a vertical orientation with its lower end

A spring with spring constant k=620 N / m is placed in a...

A spring with spring constant $k=620 \mathrm{N} / \mathrm{m}$ is placed in a vertical orientation with its lower end supported by a horizontal surface. The upper end is depressed $25 \mathrm{cm},$ and a block with a weight of 50 $\mathrm{N}$ is placed (unattached) on the depressed spring. The system is then released from rest. Assume that the gravitational potential energy $U_{g}$ of the block is zero at the release point $(y=0)$ and calculate the kinetic energy $K$ of the block for $y$ equal to (a) 0 , (b) $0.050 \mathrm{m},(\mathrm{c}) 0.10 \mathrm{m},(\mathrm{d}) 0.15 \mathrm{m},$ and $(\mathrm{e}) 0.20 \mathrm{m} .$ Also, (f) how far above its point of release does the block rise?

Key Concepts

Conservation of Energy

In mechanics, conservation of energy is the principle that the total energy in a closed system remains constant. In this context, it implies that the sum of gravitational potential energy, elastic (spring) potential energy, and kinetic energy remains constant throughout the motion, allowing us to analyze how energy transforms from one form to another as the spring-block system moves.

Elastic Potential Energy

Elastic potential energy is stored in a deformable object such as a spring when it is compressed or stretched from its equilibrium position. It is typically calculated using the formula ½kx², where k is the spring constant and x is the displacement from the equilibrium position. This concept is crucial in problems involving springs since it quantifies the energy stored due to the spring’s deformation.

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. It is generally defined as U = mgy, where m is the mass, g is the gravitational acceleration, and y is the height relative to a chosen reference level. This concept helps in understanding how the height of the block affects its energy state in a vertical spring-mass system.

Kinetic Energy

Kinetic energy is the energy that a body possesses due to its motion, given by the formula K = ½mv², where m is the mass and v is the velocity of the object. It represents the energy associated with the block's movement, and in the context of energy conservation, variations in kinetic energy reflect the conversion of stored elastic and gravitational potential energy as the system evolves.

Equilibrium Position Analysis

The equilibrium position in a spring-mass system occurs where the net force on the block is zero, typically where the upward spring force equals the downward gravitational force. Analyzing the equilibrium position is essential to understand the points at which the system may momentarily have zero velocity or change direction, and this informs the analysis of oscillatory behavior after the system is released.

Transcript

00:01 Here we know that the spring constant k is going to be equal to 620 newtons per meter.

00:13 We know that the weight of the block is going to be equal to 50 newtons, which will equal essentially mg.

00:22 And we know that y initial is equal in 0 .25 meters.

00:28 So here we can say that the conservation of energy simply leads to the equation, kinetic energy initial plus the potential energy initial equals the final kinetic energy plus the final potential energy we can say that there is no kinetic energy initial and we can then say that this would be one -half k y initial squared equaling one -half mv final squared plus m gy -f plus 1 1 1� multiplied by y final minus y initial quantity squared and at this point we can say that here y initial it would be equal to the initial depression of the spring and then y final minus y initial is the displacement of the spring from its equilibrium position when the block is at y final so we can then say that the kinetic energy of the block final would be equal to 1 .5 mv final squared and this would be essentially equaling a new a new line this would be equal to one half k multiplied by y initial squared minus y final minus y initial quantity squared minus m gy final and we can say that for y final equaling zero, the final kinetic energy is zero.

02:26 So we can say at y final equaling zero, we can of course say that here the final kinetic energy is equaling zero joules.

02:37 This would be our answer for part a because of course this corresponds to the initial release point.

02:44 And initially we don't have any kinetic energy.

02:47 So with a y final equaling zero, again we have a final kinetic energy of zero.

02:52 Again, we have a final which makes sense.

02:56 For part b, we know that for a y final equaling 0 .050 meters, we have that k sub f would be equal to one -half k times again, y initial squared minus y final minus y initial quantity squared minus mgyf.

03:18 And so we can solve.

03:19 So kinetic energy final would be equal to one -half times 620, newtons per meter.

03:26 In fact, just a safe space, we're not going to write all the units down.

03:30 We can just say that this is going to be multiplied by 0 .250 squared minus 0 .050 minus 0 .250 quantity squared minus and then this would be essentially 50 newtons.

03:52 The weight multiplied by 0 .050 and we have a final kinetic energy for part b equaling 4 .48 joules.

04:04 For part c, y final now is equaling 0 .1 .00 meters and we can say that here, k final is going to be the exact same formula that we used for part b.

04:20 However, y final is now equaling 0 .1 meters.

04:23 So again, this.

04:24 This would be 1 1⁄2 times 620 multiplied by 0 .250 squared minus 0 .10 minus 0 .250 quantity squared...

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