/*! 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 67 A spring \((k=200 \mathrm{~N} / ... [FREE SOLUTION] | 91Ó°ÊÓ

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A spring \((k=200 \mathrm{~N} / \mathrm{m})\) is fixed at the top of a frictionless plane inclined at angle \(\theta=40^{\circ}\) (Fig. 8-57). A \(1.0 \mathrm{~kg}\) block is projected up the plane, from an initial position that is distance \(d=0.60 \mathrm{~m}\) from the end of the relaxed spring, with an initial kinetic energy of \(16 \mathrm{~J}\). (a) What is the kinetic energy of the block at the instant it has compressed the spring \(0.20 \mathrm{~m}\) ? (b) With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by \(0.40 \mathrm{~m}\) ?

Short Answer

Expert verified
(a) 8.22 J, (b) 18.52 J

Step by step solution

01

Determine Initial and Final Energies for Part (a)

Given initial kinetic energy \( K_i = 16 \text{ J} \) and no initial elastic potential energy, the total initial energy \( E_i = K_i + U_{i} = 16 \text{ J} \). When the block compresses the spring by \( x = 0.20 \text{ m} \), the spring's potential energy \( U = \frac{1}{2} k x^2 \), where \( k = 200 \text{ N/m} \). Substitute to find \( U = \frac{1}{2} \times 200 \times (0.20)^2 = 4 \text{ J} \).
02

Calculate Gravitational Potential Energy Change for Part (a)

The block rises along the incline height \( h \) when the spring compresses. Use \( h = x \sin \theta \) where \( x = 0.60 \text{ m} \). Thus, \( h = 0.60 \times \sin 40^\circ \approx 0.386 \text{ m} \). The change in gravitational potential energy \( \Delta U_g = mgh = 1.0 \times 9.8 \times 0.386 = 3.78 \text{ J} \).
03

Calculate Final Kinetic Energy for Part (a)

At compression \( x = 0.20 \text{ m} \), total final energy \( E_f = K_f + U + U_g \). Using energy conservation \( E_i = E_f \), we have \( 16 = K_f + 4 + 3.78 \). Solving for \( K_f \), we find \( K_f = 16 - 7.78 = 8.22 \text{ J} \).
04

Determine Required Initial Energy for Part (b)

When the block compresses the spring by \( x = 0.40 \text{ m} \) and stops, all initial kinetic energy is converted to spring and gravitational potential energies. The spring energy \( U = \frac{1}{2} \times 200 \times (0.40)^2 = 16 \text{ J} \).
05

Calculate Gravitational Potential Energy Change for Part (b)

The block's height change \( h = 0.40 \sin \theta = 0.40 \times \sin 40^\circ \approx 0.257 \text{ m} \). Thus, \( \Delta U_g = 1.0 \times 9.8 \times 0.257 \approx 2.52 \text{ J} \).
06

Calculate Required Initial Kinetic Energy for Part (b)

Using conservation, the initial kinetic energy needed \( K_i + U_i + U_{g,i} = U + U_g \). With \( U_i = 0 \), we have \( K_i = 16 + 2.52 = 18.52 \text{ J} \).

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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 that an object possesses due to its motion. In the context of physics, it is an important aspect when analyzing systems involving moving objects, such as a block sliding up an inclined plane. The formula for kinetic energy (\( K \)) is:\[K = \frac{1}{2}mv^2\] where \( m \) is the mass of the object, and \( v \) is the velocity. As seen in the original problem, before the block moves up the incline and compresses the spring, it had an initial kinetic energy of 16 J. This energy was crucial in enabling the block to ascend the plane and perform work against gravitational forces and spring compression.
  • Kinetic energy is directly proportional to the mass and the square of the velocity of the object.
  • The more kinetic energy an object has, the faster it moves.
  • Kinetic energy can be transformed into other types of energy during interactions, such as potential energy.
Understanding kinetic energy allows us to predict how an object will behave during motion and interactions with other forces.
Potential Energy
Potential energy is stored energy that an object has due to its position or configuration. In the problem, we dealt with both gravitational potential energy and spring potential energy.
  • Gravitational Potential Energy: This is the energy that an object possesses because of its position in a gravitational field. It is calculated using the formula \( U_g = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above a reference point. As the block rises up the incline, it gains gravitational potential energy.
  • Spring Potential Energy: When dealing with springs, the potential energy is stored as the spring is compressed or stretched. The potential energy is calculated using Hooke's Law:\[U_s = \frac{1}{2}kx^2\]where \( k \) is the spring constant and \( x \) is the compression or extension length from the spring’s equilibrium position. In the exercise, this energy is a critical factor determining how the block interacts with the spring.
It is important to note that potential energy stored in the system can be converted back to kinetic energy if the conditions allow, such as when a compressed spring is released. Potential energy plays a vital role in energy conservation and transformation across different forms.
Spring Constant
The spring constant, denoted by \( k \), is a measure of the stiffness or rigidity of a spring. It quantifies the force required to compress or stretch the spring by a unit length. According to Hooke’s Law, the force exerted by a spring is directly proportional to the displacement, and is given by:\[F = -kx\]where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position. In this exercise, the given spring constant is 200 N/m, indicating that 200 N of force is needed to compress or stretch the spring by one meter.
  • Higher spring constants mean stiffer springs, requiring more force to displace.
  • A softer spring with a lower spring constant will compress more easily with less force.
  • Spring potential energy can be determined using the spring constant.
Understanding the spring constant is essential for predicting how a spring will behave when forces are applied. It helps in calculating the potential energy stored in the spring, which is crucial for systems where springs and motion are involved. The spring constant is foundational in mechanics for designing and understanding systems involving elastic materials.

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Most popular questions from this chapter

A \(700 \mathrm{~g}\) block is released from rest at height \(h_{0}\) above a vertical spring with spring constant \(k=400 \mathrm{~N} / \mathrm{m}\) and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring \(19.0 \mathrm{~cm}\). How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of \(h_{0} ?(\mathrm{~d})\) If the block were released from height \(2.00 h_{0}\) above the spring, what would be the maximum compression of the spring?

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 \(50 \mathrm{~g}\) ball is thrown from a window with an initial velocity of \(8.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) above the horizontal. Using energy methods, determine (a) the kinetic energy of the ball at the top of its flight and (b) its speed when it is \(3.0 \mathrm{~m}\) below the window. Does the answer to (b) depend on either (c) the mass of the ball or (d) the initial angle?

A block of mass \(m=2.0 \mathrm{~kg}\) is dropped from height \(h=40 \mathrm{~cm}\) onto a spring of spring constant \(k=1960 \mathrm{~N} / \mathrm{m}\) (Fig. \(8-37\) ). Find the maximum distance the spring is compressed.

A \(5.0 \mathrm{~kg}\) block is projected at \(5.0 \mathrm{~m} / \mathrm{s}\) up a plane that is inclined at \(30^{\circ}\) with the horizontal. How far up along the plane does the block go (a) if the plane is frictionless and (b) if the coefficient of kinetic friction between the block and the plane is \(0.40 ?\) (c) In the latter case, what is the increase in thermal energy of block and plane during the block's ascent? (d) If the block then slides back down against the frictional force, what is the block's speed when it reaches the original projection point?

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