/*! 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 68 cant copy graph A box of mass ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 68

cant copy graph A box of mass \(m\) is released from rest at the top of a frictionless ramp tilted at an angle \(\theta\) from the horizontal. The box slides down a distance \(s\) and collides with spring with a spring constant \(k\). The spring then compresses a distance \(x\). To find \(x\) one should (A) equate \(m g s\) to \(1 / 2 k^2\) and solve for \(x\). (B) equate \(m g s \cos \theta\) to \(1 / 2 k x^2\) and solve for \(x\). (C) equate \(m g s \sin \theta\) to \(1 / 2 k x^2\) and solve for \(x\). (D) equate \(m g(x+s) \sin \theta\) to \(1 / 2 k x^2\) and solve for \(x\). (E) equate \(m g\) to \(k x\) and solve for \(x\).

Short Answer

Expert verified
Answer: The correct equation is (C) \(mgs\sin \theta = \frac{1}{2}kx^2\), where \(m\) is the mass of the box, \(g\) is the acceleration due to gravity, \(s\) is the distance along the ramp, \(\theta\) is the angle of inclination of the ramp, \(k\) is the spring constant, and \(x\) is the compression of the spring.

Step by step solution

01

State work-energy theorem

The work-energy theorem states that the work done on an object is equal to its change in kinetic energy: \(W = \Delta KE\)
02

Apply work-energy theorem to the box

When the box is released from the top of the ramp and slides down a distance \(s\), it loses gravitational potential energy and gains kinetic energy. The work-energy theorem can be written as \(mgh = \frac{1}{2} mv^2\), where \(h\) is the vertical height and \(v\) is the final velocity of the box. We can rewrite this, taking into account the \(\theta\) and \(s\), \(mgs\sin \theta = \frac{1}{2} mv^2\), where \(v\) is the velocity at the bottom of the ramp.
03

Apply work-energy theorem to the spring

When the box collides with the spring, its kinetic energy is converted into potential energy stored in the spring. The work-energy theorem for the spring can be written as \(\frac{1}{2}mv^2 = \frac{1}{2}kx^2\).
04

Combine equations from Steps 2 and 3

Combining the equations from steps 2 and 3, we have the equation: \(mgs\sin \theta = \frac{1}{2}kx^2\).
05

Compare the combined equation with the options given

Comparing the combined equation with the given options, we see that option (C) equating \(m g s \sin \theta\) to \(\frac{1}{2} k x^2\) and solving for \(x\), matches our conclusion. Therefore, the correct answer is (C).

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Key Concepts

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

Frictionless Ramp
A frictionless ramp is a slope on which an object can slide without the resistance we usually know as friction. This means that when a box is released from the top of such a ramp, it won't slow down due to any resistance from the surface.

This allows us to focus only on other forces, such as gravity, that act on the box.
  • The angle of the ramp, denoted by \(\theta\), plays a crucial role in determining how efficiently gravity can pull the box down.
  • The steeper the ramp, or the larger the \(\theta\), the faster the box will pick up speed as it slides down.
  • Without friction, the only force doing work on the box is gravity, which simplifies the calculations significantly.
Understanding the concept of a frictionless ramp is essential in applying the work-energy theorem correctly in problems like the one in question.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy that an object possesses due to its position in a gravitational field. For an object like our box on the ramp, this energy is dependent on its height above the ground.

The formula for gravitational potential energy is given by \(U = mgh\), where:
  • \(m\) is the mass of the object,
  • \(g\) is the acceleration due to gravity, approximately \(9.8 \, \text{m/s}^2\) on Earth,
  • \(h\) is the height of the object above a reference point.
In the scenario with the ramp, the vertical component of the slide contributes to the change in gravitational potential energy as the box slides down and converts this energy into kinetic energy. This is expressed as \(mgs\sin \theta\), where \(s\) is the distance along the ramp.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. When the box starts sliding down the ramp, the gravitational potential energy it had at the top gets converted into kinetic energy.

The formula for kinetic energy is \(KE = \frac{1}{2}mv^2\), where:
  • \(m\) is the mass of the object, and
  • \(v\) is the velocity of the object.
As the box slides down the ramp, it speeds up, which means its kinetic energy increases. By the time the box reaches the bottom of the ramp and hits the spring, all of its initial gravitational potential energy has been converted into kinetic energy, assuming no energy loss due to friction.
Spring Potential Energy
Spring potential energy is the energy stored in a compressed or stretched spring. When the moving box collides with the spring, it starts compressing it.

The energy stored in the spring, which is potential energy, can be calculated using the formula \(PE = \frac{1}{2}kx^2\), where:
  • \(k\) is the spring constant, measuring how stiff the spring is, and
  • \(x\) is the displacement of the spring from its equilibrium position.
The box compresses the spring until all of its kinetic energy has been transferred into spring potential energy, which is calculated as \(\frac{1}{2}kx^2\) in this context. This reflects the point where the box stops moving, having converted all of its energy into compressing the spring.

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Most popular questions from this chapter

Model rocket engines burn for only a short time before using up all their propellant. Suppose a model rocket is launched from its stand at an angle \(60^{\circ}\) above the horizontal. The mass of the rocket is \(0.25 \mathrm{~kg}\) and its final speed is \(50 \mathrm{~m} / \mathrm{s}\). If the engine burns for \(1.25 \mathrm{~s}\), the impulse it gives to the rocket is most nearly (A) \(15.6 \mathrm{~N} \mathrm{~s}\) (B) \(12.5 \mathrm{~N} \mathrm{~s}\) (C) \(6.25 \mathrm{~N} \mathrm{~s}\) (D) \(\frac{\sqrt{3}}{2} \times 12.5 \mathrm{~N} \mathrm{~s}\) (E) \(12.5 \mathrm{~N}\)

Two planets, Gamma and Delta, with masses \(m_G\) and \(m_D\) and radii \(R_G\) and \(R_D\) are initially at rest a large distance \(d\) apart from each other. Khan the Conqueror suddenly allows the two planets to approach under their mutual gravitational attraction until they collide. To determine the velocity at which Gamma and Delta collide you could (A) use conservation of energy alone. (B) use conservation of momentum alone. (C) use both conservation of energy and momentum. (D) The problem cannot be solved using energy and momentum conservation. (E) The problem cannot be solved unless you know how long it takes the two planets to collide.

cant copy graph Two students, Alice and Bob, decide to compute the power that the Earth's gravitational field expends on a block of mass \(m\) as the block slides down a frictionless inclined plane. Alice reasons: "The gravitational force pulling the block down the incline is \(F=m g \sin \theta\). The block's velocity at any given height \(h\) from the top of the incline is \(v=\sqrt{2 g h}\). Power is defined as force \(\times\) velocity. Therefore, the power is \(P=m g \sin \theta \sqrt{2 g h}\)." Bob reasons: "Power is \(\Delta W / \Delta t\). By the work-energy theorem, the change in work is the change in kinetic energy, but without friction \(\Delta W=m g h\). The change in time is \(\Delta t=\Delta v / a, \Delta v=\sqrt{2 g h}\) and \(a=g \sin \theta\). So \(\Delta t=\sqrt{2 g h} /(g \sin \theta)\). Therefore, \(P=m g h \times \frac{g \sin \theta}{\sqrt{2 g h}}=m g \sin \theta \sqrt{\frac{g h}{2}}\)." Alice and Bob look at each other and scratch their heads. Who is correct? \(\begin{array}{llll}\text { i. Alice } & \text { ii. Bob } & \text { iii. Neither iv. Both } & \text { v. The problem is imprecisely worded. }\end{array}\) (A) \(\mathrm{i}\) (B) ii (C) iii (D) iv (E) iv and \(v\)

Two masses, \(m_1\) and \(m_2\), are connected by a string of negligible mass, which passes over a massless pulley, as shown in the figure. Assume \(m_1>m_2\). What is the acceleration \(a\) of \(m_1\) ? (A) \(a=\left(m_1-m_2\right) g /\left(m_1+m_2\right)\) (B) \(a=\left(m_1+m_2\right) g /\left(m_1-m_2\right)\) (C) \(a=\left(m_1 m_2\right) g /\left(m_1+m_2\right)\) (D) \(a=\left(m_2-m_1\right) g /\left(m_1+m_2\right)\) (E) \(a=\left(m_1-m_2\right) g /\left(m_1 m_2\right)\)

Two objects, \(m\) and \(M\), with initial velocities \(v_0\) and \(U_0\) respectively, undergo a one-dimensional elastic collision. Show that their relative speed is unchanged during the collision. (The relative velocity between \(m\) and \(M\) is defined as \(v-U_{-}\))
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