/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 130 A spring with spring constant \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A spring with spring constant \(k=200 \mathrm{~N} / \mathrm{m}\) is suspended vertically with its upper end fixed to the ceiling and its lower end at position \(y=0 .\) A block of weight \(20 \mathrm{~N}\) is attached to the lower end, held still for a moment, and then released. What are (a) the kinetic energy \(K,(\mathrm{~b})\) the change (from the initial value) in the gravitational potential energy \(\Delta U_{g}\), and (c) the change in the elastic potential energy \(\Delta U_{e}\) of the spring-block system when the block is at \(y=-5.0 \mathrm{~cm}\) ? What are (d) \(K\), (e) \(\Delta U_{g}\), and (f) \(\Delta U_{e}\) when \(y=-10 \mathrm{~cm},(\mathrm{~g}) K,(\mathrm{~h}) \Delta U_{g}\), and \((\mathrm{i}) \Delta U_{e}\) when \(y=-15 \mathrm{~cm}\), and \((\mathrm{j}) K,(\mathrm{k}) \Delta U_{g}\), and (1) \(\Delta U_{e}\) when \(y=-20 \mathrm{~cm}\) ?

Short Answer

Expert verified
(a) 0.75 J, (b) 1 J, (c) 0.25 J at -5 cm; (d) 1 J, (e) 2 J, (f) 1 J at -10 cm; (g) 0.75 J, (h) 3 J, (i) 2.25 J at -15 cm; (j) 0 J, (k) 4 J, (l) 4 J at -20 cm.

Step by step solution

01

Identify Known and Unknowns

We are given that the spring constant is \(k = 200\, \text{N/m}\) and the block's weight is \(20\, \text{N}\). The initial position is \(y = 0\). We need to find kinetic energy \(K\), change in gravitational potential energy \(\Delta U_g\), and change in elastic potential energy \(\Delta U_e\) at different positions (\(y = -5\, \text{cm}\), \(-10\, \text{cm}\), \(-15\, \text{cm}\), \(-20\, \text{cm}\)).
02

Formula for Potential Energies

Gravitational potential energy change is \(\Delta U_g = mg\Delta y\), where \(m = \frac{W}{g}\) is the mass and \(W = 20\, \text{N}\). Elastic potential energy change is \(\Delta U_e = \frac{1}{2}k(\Delta y)^2\), where \(k = 200\, \text{N/m}\).
03

Calculate Motion and Energy at \(y = -5\, \text{cm}\)

Find mass: \(m = \frac{20 \text{ N}}{9.8 \text{ m/s}^2} \approx 2.04 \text{ kg}\). \(\Delta U_g = 20 \cdot (-0.05) = -1 \text{ J}\). \(\Delta U_e = \frac{1}{2} \cdot 200 \cdot (-0.05)^2 = 0.25 \text{ J}\). Again using conservation of energy, \(K = - (\Delta U_g + \Delta U_e) = 0.75 \text{ J}\).
04

Calculate Motion and Energy at \(y = -10\, \text{cm}\)

\(\Delta U_g = 20 \cdot (-0.10) = -2 \text{ J}\). \(\Delta U_e = \frac{1}{2} \cdot 200 \cdot (-0.10)^2 = 1 \text{ J}\). \(K = -(\Delta U_g + \Delta U_e) = 1 \text{ J}\).
05

Calculate Motion and Energy at \(y = -15\, \text{cm}\)

\(\Delta U_g = 20 \cdot (-0.15) = -3 \text{ J}\). \(\Delta U_e = \frac{1}{2} \cdot 200 \cdot (-0.15)^2 = 2.25 \text{ J}\). \(K = -(\Delta U_g + \Delta U_e) = 0.75 \text{ J}\).
06

Calculate Motion and Energy at \(y = -20\, \text{cm}\)

\(\Delta U_g = 20 \cdot (-0.20) = -4 \text{ J}\). \(\Delta U_e = \frac{1}{2} \cdot 200 \cdot (-0.20)^2 = 4 \text{ J}\). \(K = -(\Delta U_g + \Delta U_e) = 0 \text{ J}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When a block is released from rest on a spring, its kinetic energy increases as it moves downward. The block begins to accelerate due to gravity and the force exerted by the spring. The equation for kinetic energy is given by: \[\ K = \frac{1}{2}mv^2\ \] where \( m \) is the mass and \( v \) is the velocity of the block.
As the block oscillates on the spring, its kinetic energy fluctuates. At its release point, \( y = 0 \), the kinetic energy is zero because the block hasn’t moved yet. When the block reaches a lower point, like \( y = -5 \text{ cm} \), it speeds up, increasing its kinetic energy.
Kinetic energy changes throughout the motion, reaching its maximum when the gravitational pull and elastic force combine optimally. Tracking kinetic energy helps us understand how energy transforms from potential forms (gravitational and elastic) into motion.
Gravitational Potential Energy
Gravitational potential energy is the energy held by an object because of its position relative to a gravitational field. For our block and spring system, it is determined by the height of the block above its equilibrium position. The equation is:\[ \Delta U_g = mg \Delta y \],
where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( \Delta y \) is the change in height.
When the block is attached to the spring at position \( y = 0 \), it is at its maximum height during the motion cycle, so its gravitational potential energy is at the maximum relative to lower positions. When the block moves downwards to positions like \( y = -5 \text{ cm} \), its gravitational potential energy decreases.
Elastic Potential Energy
Elastic potential energy is stored in spring systems when the spring is either compressed or stretched from its natural length. The energy stored in a spring is described by Hooke’s law:
\[ \Delta U_e = \frac{1}{2}k(\Delta y)^2\]
Here, \( k \) is the spring constant, and \( \Delta y \) is the displacement from its unstressed position.
As the block descends and stretches the spring, elastic potential energy accumulates. For instance, at \( y = -5 \text{ cm} \), the elastic energy starts increasing because the spring is stretched. This energy can be transformed back into kinetic energy and gravitational potential energy, showcasing the interplay of forces and energy conservation in spring-block systems.

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Most popular questions from this chapter

A \(60.0 \mathrm{~kg}\) circus performer slides \(4.00 \mathrm{~m}\) down a pole to the circus floor, starting from rest. What is the kinetic energy of the performer as she reaches the floor if the frictional force on her from the pole (a) is negligible (she will be hurt) and (b) has a magnitude of \(500 \mathrm{~N}\) ?

A \(3.2 \mathrm{~kg}\) sloth hangs \(3.0 \mathrm{~m}\) above the ground. (a) What is the gravitational potential energy of the sloth-Earth system if we take the reference point \(y=0\) to be at the ground? If the sloth drops to the ground and air drag on it is assumed to be negligible, what are the (b) kinetic energy and (c) speed of the sloth just before it reaches the ground?

A \(2.50 \mathrm{~kg}\) beverage can is thrown directly downward from a height of \(4.00 \mathrm{~m}\), with an initial speed of \(3.00 \mathrm{~m} / \mathrm{s}\). The air drag on the can is negligible. What is the kinetic energy of the can (a) as it reaches the ground at the end of its fall and (b) when it is halfway to the ground? What are (c) the kinetic energy of the can and (d) the gravitational potential energy of the can-Earth system \(0.200 \mathrm{~s}\) before the can reaches the ground? For the latter, take the reference point \(y=0\) to be at the ground.

A stone with a weight of \(5.29 \mathrm{~N}\) is launched vertically from ground level with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s}\), and the air drag on it is \(0.265 \mathrm{~N}\) throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground?

A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant \(k=400 \mathrm{~N} / \mathrm{m} ;\) the other end of the spring is fixed in place. The cookie has a kinetic energy of \(20.0 \mathrm{~J}\) as it passes through the spring's equilibrium position. As the cookie slides, a frictional force of magnitude \(10.0 \mathrm{~N}\) acts on it. (a) How far will the cookie slide from the equilibrium position before coming momentarily to rest? (b) What will be the kinetic energy of the cookie as it slides back through the equilibrium position?

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