/*! 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) 5.47 m/s, (b) 5.38 m, (c) Remain the same.

Step by step solution

01

Calculate Gravitational Force Component Down Incline

The gravitational force component parallel to the ramp is \[ F_{\text{gravity, parallel}} = mg \sin \theta. \]Given that the mass \( m = 180 \) kg, \( g = 9.8 \) m/s², and the angle \( \theta = 39^{\circ} \), calculate:\[ F_{\text{gravity, parallel}} = 180 \times 9.8 \times \sin(39^{\circ}) \approx 1110.83 \text{ N}. \]
02

Calculate Frictional Force on the Ramp

The frictional force on the ramp is calculated as:\[ F_{\text{friction}} = \mu_k \cdot N = \mu_k \cdot mg \cos \theta. \]Where \( \mu_k = 0.28 \) is the coefficient of friction and \( N = mg \cos \theta \) is the normal force. Calculate:\[ F_{\text{friction}} = 0.28 \times 180 \times 9.8 \times \cos(39^{\circ}) \approx 380.85 \text{ N}. \]
03

Calculate Net Force and Work Done on the Ramp

The net force acting on the crate along the ramp is:\[ F_{\text{net}} = F_{\text{gravity, parallel}} - F_{\text{friction}} = 1110.83 - 380.85 \approx 729.98 \text{ N}. \]The work done by the net force over the distance of the ramp \( d = 3.7 \) m is:\[ W = F_{\text{net}} \cdot d = 729.98 \times 3.7 \approx 2700.93 \text{ J}. \]
04

Use Work-Energy Theorem to Find Final Velocity

The work-energy theorem states that the work done is equal to the change in kinetic energy:\[ W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2. \]Since the initial velocity \( u = 0 \), the equation simplifies to:\[ 2700.93 = \frac{1}{2} \times 180 \times v^2. \]Solving for \( v \), we find:\[ v = \sqrt{\frac{2700.93 \times 2}{180}} \approx 5.47 \text{ m/s}. \]
05

Calculate Distance on Horizontal Floor

The crate slides across the horizontal floor with the kinetic friction acting as the only horizontal force. Using the initial velocity \( v = 5.47 \) m/s and noting that the work done by friction \( W_{ ext{floor}} \) will equal the kinetic energy at the bottom:\[ W_{\text{floor}} = \mu_k mg \cdot d_{\text{floor}} = \frac{1}{2}mv^2. \]Solve for the distance:\[ 0.28 \cdot 180 \cdot 9.8 \cdot d_{\text{floor}} = \frac{1}{2} \times 180 \times 5.47^2, \]\[ d_{\text{floor}} = \frac{0.5 \times 180 \times 5.47^2}{0.28 \times 180 \times 9.8} \approx 5.38 \text{ m}. \]
06

Analyze Effect of Halving Mass

If the mass is halved, both the gravitational force component and frictional force scale linearly with mass, which means they cancel each other out in how they affect acceleration and net force. Since the coefficients remain unchanged, the speed at the bottom and distance on the floor remain the same.

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Key Concepts

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

Inclined Plane
Inclined planes might seem like simple structures, but they play a crucial role in physics problems. Consider a ramp, in this scenario, it's 3.7 meters long set at an angle of 39 degrees. The steeper the angle, the more component of the gravitational force acts along the plane, causing objects to accelerate downwards more significantly.

When analyzing a crate sliding down a ramp, it is vital to determine the gravitational force component that acts along the incline. The force can be calculated using the formula \( F_{\text{gravity, parallel}} = mg \sin \theta \), where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the incline. This force is responsible for pulling the crate downwards along the slope.

Working with inclined planes often involves calculating the opposing forces like friction. The net force, which determines how fast the crate can slide down, is the difference between the gravitational component and the frictional force acting against this motion.
Kinetic Friction
Kinetic friction is a force that acts against the motion of sliding objects. It plays a crucial role when an object, like our crate, transitions from a state of rest to sliding down a ramp.

The magnitude of kinetic friction is determined by the coefficient of kinetic friction, \( \mu_k \), and the normal force, \( N \). For inclined planes, the normal force differs because it's perpendicular to the surface of the incline and is calculated using \( N = mg \cos \theta \). Together, they dictate the frictional force as \( F_{\text{friction}} = \mu_k N \).

As the crate slides down, friction will oppose the motion, working to slow it by converting kinetic energy into heat. It’s essential to calculate this force to understand how much it counters the gravitational pull and affects the speed of the sliding object.
  • High kinetic friction means more opposition to motion.
  • The coefficient of friction depends on the surfaces in contact.
  • Friction reduces the net force causing less acceleration in downwards motion.
Work-Energy Theorem
The work-energy theorem is a fundamental concept stating that the work done on an object is equal to its change in kinetic energy. This principle helps us understand how energy gets transferred and transformed when the crate moves down the ramp and across the floor.

In this scenario, the work done by the net force during the crate's descent represents the energy change from gravitational potential to kinetic energy. Mathematically, this is expressed as \( W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \), where \( W \) is the work, \( m \) is the mass and \( v \) is the final velocity.

Initially, the crate was at rest, making its initial kinetic energy zero. As the crate travels down the slope, energy initially stored as potential energy transforms into kinetic energy, increasing its velocity. Understanding this theorem helps us make similar calculations when dealing with energy conservation in other systems.

Ultimately, this fundamental theorem explains not only how energy changes but also helps us predict the velocity acquired by objects under the influence of forces—an essential tool for solving many physics problems.

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Most popular questions from this chapter

If a \(70 \mathrm{~kg}\) baseball player steals home by sliding into the plate with an initial speed of \(10 \mathrm{~m} / \mathrm{s}\) just as he hits the ground, (a) what is the decrease in the player's kinetic energy and (b) what is the increase in the thermal energy of his body and the ground along which he slides?

We move a particle along an \(x\) axis, first outward from \(x=1.0 \mathrm{~m}\) to \(x=4.0 \mathrm{~m}\) and then back to \(x=1.0 \mathrm{~m}\), while an external force acts on it. That force is directed along the \(x\) axis, and its \(x\) component can have different values for the outward trip and for the return trip. Here are the values (in newtons) for four situations, where \(x\) is in meters: $$ \begin{array}{ll} \hline \text { Outward } & \text { Inward } \\ \hline \text { (a) }+3.0 & -3.0 \\ \text { (b) }+5.0 & +5.0 \\ \text { (c) }+2.0 x & -2.0 x \\ \text { (d) }+3.0 x^{2} & +3.0 x^{2} \\ \hline \end{array} $$ Find the net work done on the particle by the external force for the round trip for each of the four situations. (e) For which, if any, is the external force conservative?

Resistance to the motion of an automobile consists of road friction, which is almost independent of speed, and air drag, which is proportional to speed- squared. For a certain car with a weight of \(12000 \mathrm{~N}\), the total resistant force \(F\) is given by \(F=300+1.8 v^{2}\), with \(F\) in newtons and \(v\) in meters per second. Calculate the power (in horsepower) required to accelerate the car at \(0.92 \mathrm{~m} / \mathrm{s}^{2}\) when the speed is \(80 \mathrm{~km} / \mathrm{h}\).

To make a pendulum, a \(300 \mathrm{~g}\) ball is attached to one end of a string that has a length of \(1.4 \mathrm{~m}\) and negligible mass. (The other end of the string is fixed.) The ball is pulled to one side until the string makes an angle of \(30.0^{\circ}\) with the vertical; then (with the string taut) the ball is released from rest. Find (a) the speed of the ball when the string makes an angle of \(20.0^{\circ}\) with the vertical and (b) the maximum speed of the ball. (c) What is the angle between the string and the vertical when the speed of the ball is one-third its maximum value?

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?

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