/*! 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 95 A factory worker accidentally re... [FREE SOLUTION] | 91Ó°ÊÓ

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A factory worker accidentally releases a \(180 \mathrm{~kg}\) crate that was being held at rest at the top of a ramp that is \(3.7 \mathrm{~m}\) long and inclined at \(39^{\circ}\) to the horizontal. The coefficient of kinetic friction between the crate and the ramp, and between the crate and the horizontal factory floor, is \(0.28 .\) (a) How fast is the crate moving as it reaches the bottom of the ramp? (b) How far will it subsequently slide across the floor? (Assume that the crate's kinetic energy does not change as it moves from the ramp onto the floor.) (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we halve the mass of the crate?

Short Answer

Expert verified
(a) Speed from ramp is calculated using net acceleration. (b) Slide distance depends on kinetic energy and friction. (c) Both are unaffected by mass change.

Step by step solution

01

Understand the Problem

We need to find the speed of a crate sliding down an inclined ramp and how far it slides on a flat surface before stopping. The ramp is incline at 39° with friction, and the entire crate's kinetic energy transfers to the flat surface.
02

Analyze Forces and Motion on the Ramp

As the crate slides down, it is influenced by gravitational force, normal force, and frictional force. The gravitational force along the incline is given by \(mg \sin \theta\), and the normal force is \(mg \cos \theta\). The frictional force is \(\mu_k \cdot N = \mu_k \cdot mg \cos \theta\). Calculate each: - Gravitational force component along the ramp: \[ F_g = mg \sin 39^{\circ} \]- Normal force: \[ N = mg \cos 39^{\circ} \]- Frictional force: \[ F_f = 0.28 \cdot mg \cos 39^{\circ} \]
03

Calculate Net Force and Acceleration on the Ramp

The net force along the ramp is the difference between gravitational force down the slope and the frictional force. Calculate net force and use it to find acceleration:- Net force: \[ F_{net} = mg \sin 39^{\circ} - 0.28 \cdot mg \cos 39^{\circ} \] - Acceleration \(a\) of the crate: \[ a = \frac{F_{net}}{m} = g \sin 39^{\circ} - 0.28g \cos 39^{\circ} \]
04

Compute Final Speed at the Bottom of the Ramp

Using the formula with initial velocity \(v_i = 0\) and displacement \(d = 3.7\, \text{m}\): \[ v^2 = v_i^2 + 2a d \] Where \(v_i = 0\), so: \[ v = \sqrt{2 \cdot a \cdot 3.7} \]
05

Calculate Distance Slid Across the Floor

The kinetic energy at the bottom of the ramp is converted to work done against friction on the floor until stop. Initial kinetic energy: \[ KE = \frac{1}{2}mv^2 \]The work done by friction is \[ \mu_k mg d \], where \(d\) is the distance on the flat floor. Equate these: \[ \mu_k mg d = \frac{1}{2}mv^2 \]Solve for \(d\):\[ d = \frac{v^2}{2\mu_k g} \]
06

Consider Effects of Mass Change

The calculated speed \(v\) and slide distance \(d\) are independent of mass \(m\) (since \(m\) cancels out in the equations). Therefore, halving the crate's mass does not affect either the speed at the bottom or the distance slid.

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Key Concepts

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

Understanding Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When the factory worker releases the crate on the ramp, gravitational potential energy converts into kinetic energy. The formula for kinetic energy is given by \[ KE = \frac{1}{2}mv^2 \] where \(m\) is the mass and \(v\) is the velocity of the object. The key idea is that as the crate slides down, this energy increases because its velocity increases. This change in form of energy allows us to analyze how fast the crate is moving.
  • Kinetic energy depends on both mass and velocity.
  • In energy conservation, potential energy converts entirely into kinetic energy minus any energy lost to other forces like friction.
  • In this problem, kinetic energy plays a crucial role in predicting how far the crate slides across the flat surface.
Break down complex formulas into comprehensible parts. Understand that velocity affects kinetic energy quadratically, meaning a small change in speed has a significant impact on its energy. This is why understanding how kinetic energy works helps to solve physics problems efficiently.
The Role of Kinetic Friction
Kinetic friction is a force that opposes the motion of an object sliding on a surface. It's different from static friction, which prevents an object from starting to move. Here, kinetic friction acts both when the crate slides down the inclined plane and when it travels across the horizontal floor. On the ramp, the frictional force can be calculated as:\[ F_f = \mu_k \cdot N = \mu_k \cdot mg \cos \theta \]where \(\mu_k\) is the coefficient of kinetic friction, \(N\) is the normal force, and \(\theta\) is the angle of the incline. This formula shows how friction opposes the component of gravity pulling the object down the slope. Key points about kinetic friction:
  • It acts against the direction of motion.
  • The coefficient of kinetic friction, \(\mu_k\), is a dimensionless value representing the interaction between the surfaces.
  • Kinetic frictional force is constant for a given velocity and normal force, simplifying calculations in physics problems involving constant speeds.
By understanding kinetic friction, we can accurately calculate how energy dissipates from the system, affecting the motion and final position of the crate.
Inclined Planes Made Easy
Inclined planes are surfaces set at an angle to the horizontal. In physics problems, they provide a classic setup for understanding motion and forces. When dealing with inclined planes, we must analyze how forces break into components along and perpendicular to the plane's surface.For a crate sliding down an inclined plane:
  • The gravitational force component parallel to the plane is \(mg \sin \theta\).
  • The normal force, which is perpendicular, is \(mg \cos \theta\).
  • These components help determine net forces and subsequent accelerations.
Effects of inclined planesIf the ramp angle increases, the gravitational component increases, accelerating the object more quickly. Conversely, a more gradual incline reduces this component, slowing the acceleration.Understanding inclined planes simplifies how to break down force vectors into manageable parts. It helps us solve not just sliding objects problems but extends to understanding more complex physics principles.

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Most popular questions from this chapter

131 Fasten one end of a vertical spring to a ceiling, attach a cabbage to the other end, and then slowly lower the cabbage until the upward force on it from the spring balances the gravitational force on it. Show that the loss of gravitational potential energy of the cabbage-Earth system equals twice the gain in the spring's potential energy.

A cookie jar is moving up a \(40^{\circ}\) incline. At a point \(55 \mathrm{~cm}\) from the bottom of the incline (measured along the incline), the jar has a speed of \(1.4 \mathrm{~m} / \mathrm{s}\). The coefficient of kinetic friction between jar and incline is \(0.15 .\) (a) How much farther up the incline will the jar move? (b) How fast will it be going when it has slid back to the bottom of the incline? (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we decrease the coefficient of kinetic friction (but do not change the given speed or location)?

Two blocks, of masses \(M=2.0 \mathrm{~kg}\) and \(2 M,\) are connected to a spring of spring constant \(k=200 \mathrm{~N} / \mathrm{m}\) that has one end fixed, as shown in Fig. \(8-69 .\) The horizontal surface and the pulley are frictionless, and the pulley has negligible mass. The blocks are released from rest with the spring relaxed. (a) What is the combined kinetic energy of the two blocks when the hanging block has fallen \(0.090 \mathrm{~m} ?\) (b) What is the kinetic energy of the hanging block when it has fallen that \(0.090 \mathrm{~m} ?\) (c) What maximum distance does the hanging block fall before momentarily stopping?

A block with mass \(m=2.00 \mathrm{~kg}\) is placed against a spring on a frictionless incline with angle \(\theta=30.0^{\circ}\) (Fig. 8-44). (The block is not attached to the spring.) The spring, with spring constant \(k=19.6 \mathrm{~N} / \mathrm{cm},\) is compressed \(20.0 \mathrm{~cm}\) and then released. (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block- Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?

A \(15 \mathrm{~kg}\) block is accelerated at \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) along a horizontal frictionless surface, with the speed increasing from \(10 \mathrm{~m} / \mathrm{s}\) to \(30 \mathrm{~m} / \mathrm{s} .\) What are \((\mathrm{a})\) the change in the block's mechanical energy and (b) the average rate at which energy is transferred to the block? What is the instantaneous rate of that transfer when the block's speed is (c) \(10 \mathrm{~m} / \mathrm{s}\) and (d) \(30 \mathrm{~m} / \mathrm{s} ?\)

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