A block of mass m=4.00 kg is attached to a spring (k=32.0 N / m) by a rope that hangs over a p

A block of mass m=4.00 kg is attached to a spring (k=32.0...

A block of mass $m=4.00 \mathrm{~kg}$ is attached to a spring $(k=32.0 \mathrm{~N} / \mathrm{m})$ by a rope that hangs over a pulley of $\operatorname{mass} M=8.00 \mathrm{~kg}$ and $\mathrm{ra}-$ dius $R=5.00 \mathrm{~cm},$ as shown in the figure. Treating the pulley as a solid homogeneous disk, neglecting friction at the axle of the pulley, and assuming the system starts from rest with the spring at its natural length, find (a) the speed of the block after it falls $1.00 \mathrm{~m}$, and (b) the maximum extension of the spring.

Key Concepts

Conservation of Mechanical Energy

This concept involves accounting for all forms of energy in a system and recognizing that, in the absence of non-conservative forces, the total mechanical energy remains constant. In problems involving a mass-spring system, gravitational potential energy, kinetic energy, and elastic potential energy must all be considered. Additionally, when a rotating object like a pulley is involved, its rotational kinetic energy is also incorporated into the conservation equation.

Hooke's Law

Hooke's Law describes the behavior of elastic springs by establishing that the force exerted by a spring is directly proportional to its extension or compression from its natural length. This relationship is encapsulated in the equation F = kx, where k is the spring constant and x is the displacement. This law is fundamental in determining the energy stored in the spring as elastic potential energy, which is (1/2)kx².

Rotational Inertia and Angular Kinetic Energy

Rotational inertia (or moment of inertia) quantifies an object's resistance to changes in its state of rotation. For a solid disk, the rotational inertia is given by I = (1/2)MR². When the pulley rotates, its rotational kinetic energy, calculated as (1/2)I?² where ? is the angular velocity, plays an important role in the overall energy balance of the system.

Relationship Between Linear and Angular Quantities

In systems where a rope or belt connects a rotating object to a linear moving object, the linear speed of the object is directly related to the angular speed of the rotating object through the relationship v = ?R. This connection is crucial for translating between the rotational dynamics of the pulley and the translational motion of the attached block, ensuring that energy conservation can be applied consistently across different forms of motion.

Transcript

00:01 The initial potential energy of the block would be equal to the gravitational potential energy.

00:08 We know the energy of the spring would be equalling to one -half times a spring constant, multiplied by the equilibrium position minus h quantity squared.

00:20 The equilibrium position is zero.

00:22 This would be equalling to half kh squared.

00:29 We know that the final energy would then be equalling to the sum of the, energy stored in the spring plus the rotational energy of the pulley plus the translational energy of the block.

00:42 And so we have one half kh squared plus one half i omega squared plus one half mv squared.

00:54 I would then be equaling to one half mr squared for the pulley.

01:01 And so the final kinetic the final amount of energy would then be equaling to one half kh squared plus one half multiplied by one half m r squared multiplied by v over r essentially substituting in for the angular velocity put uh having it in terms of the linear velocity quantity squared plus one half mb squared now, simplifying the final kinetic energy would then be equalling to 1 half kh squared plus v squared multiplied by the mass of the pulley divided by 4 plus the mass of the block divided by 2.

01:50 We can relate these to each other.

01:52 The initial energy is going to equal the final energy.

01:55 M .g .h would be equalling to 1 .5 kh squared plus v squared multiplied by m.

02:05 L.

02:05 Over 4 plus m over 2 and solving for a v this would be equal to the square root of m g h minus one half k h squared divided by the mass of the pulley divided by four plus the mass of the block divided by two and we find that the maximum extension of the spring will occur when the pulley has stopped rotating and the mass has stopped moving.

02:37 So we can say that this would be equalling then to one half kh squared, given that at that at the maximum extension, all of the energy of the system will be in the form of the potential energy stored in a spring...

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