/*! 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 110 A \(5.0 \mathrm{~kg}\) block is ... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
a) 2.55 m, b) 1.51 m, c) 25.6 J, d) 2.14 m/s.

Step by step solution

01

Analyze forces and energy for frictionless case

When the block is tossed upwards on a frictionless slope inclined at an angle, gravity is the only force acting in the direction of motion. The change in kinetic energy is equal to the work done by gravity. Using energy conservation,\[ \frac{1}{2} m v^2 = m g h \] where \( m = 5.0 \; \text{kg} \), \( v = 5.0 \; \text{m/s} \), and \( g = 9.8 \; \text{m/s}^2 \). Solve for \( h \), then calculate the distance \( d \) up the plane using \( h = d \sin(30^{\circ}) \).
02

Calculate height and distance for part (a)

The initial kinetic energy is \( \frac{1}{2} \times 5.0 \times (5.0)^2 = 62.5 \; \text{J} \). Set this equal to \( mgh \), thus:\[ 62.5 = 5.0 \times 9.8 \times h \]Solve for \( h \):\[ h = \frac{62.5}{5.0 \times 9.8} \approx 1.275 \; \text{m} \]Find \( d \) using:\[ d = \frac{h}{\sin(30^{\circ})} = \frac{1.275}{0.5} = 2.55 \; \text{m} \]
03

Analyze motion with friction

Now consider the kinetic friction force, \( f_k = \mu_k mg \cos(30^{\circ}) \), acting against the motion. The net work done is:\[ W = -f_k \cdot d - mgh \]and the energy equation balances:\[ \frac{1}{2} m v^2 = mgh + f_k d \]
04

Calculate distance for part (b)

Calculate the force of kinetic friction:\[ f_k = 0.40 \times 5.0 \times 9.8 \times \cos(30^{\circ}) \approx 16.97 \; \text{N} \]Using energy balance:\[ 62.5 = 5.0 \cdot 9.8 \cdot h + 16.97 \cdot d \]\( h = d \sin(30) \), substitute and solve:\[ 62.5 = 5.0 \cdot 9.8 \cdot \frac{d}{2} + 16.97 \cdot d \]Simplify and solve for \( d \):\[ 62.5 = 24.5d + 16.97d \]\[ 62.5 = 41.47d \]\[ d \approx 1.51 \; \text{m} \]
05

Calculate thermal energy increase

The increase in thermal energy equals the work done by friction:\[ \Delta E_{\text{thermal}} = f_k \cdot d = 16.97 \times 1.51 \approx 25.6 \; \text{J} \]
06

Calculate speed upon descent back to original point

When sliding back down, the block loses potential energy, which equals the sum of regained kinetic energy and work done against friction:\[ mgh = \frac{1}{2} mv^2 + f_k \cdot d \]Solve for \( v \):\[ 5.0 \cdot 9.8 \cdot \frac{1.51}{2} = \frac{1}{2} \cdot 5.0 \cdot v^2 + 25.6 \]Simplify and solve for \( v \):\[ 37.0 - 25.6 = 2.5 \cdot v^2 \]\[ v^2 = 4.56 \]\[ v \approx 2.14 \; \text{m/s} \]

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Key Concepts

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

Frictionless Motion
Imagine a scenario where a block slides up an inclined plane without any friction. In such a case, the motion can be described as frictionless. This means that the only force acting on the block, apart from its initial push, is gravity.
Gravity pulls the block downwards as it travels up the incline. The path of the block depends on its initial kinetic energy and the work done by gravity.
In physics, we use the concept of energy conservation to solve this type of problem. The equation that helps us understand frictionless motion is:
  • Initial kinetic energy = Gravitational potential energy at peak height
Using this principle, we find that all the initial kinetic energy is converted into gravitational potential energy when the block reaches the highest point on the incline. The calculation steps involve setting up the equation with known values, such as mass, initial velocity, and gravitational acceleration.
Kinetic Friction
When a moving block encounters friction, particularly kinetic friction, the motion becomes more complex than in the frictionless case.
Kinetic friction is the force that opposes the movement between two sliding surfaces. It's calculated using the formula: \[ f_k = \mu_k \cdot N \] where:
  • \( \mu_k \) is the coefficient of kinetic friction.
  • \( N \) is the normal force, which is \( mg \cdot \cos(\theta) \) on an inclined plane where \( \theta \) is the angle of the incline.
The presence of kinetic friction forces us to reconsider the energy conservation equation. In the new setup, the work done against friction plays an essential role. As the block moves uphill, the friction converts some of the block's energy into heat, reducing the distance it will travel compared to a frictionless scenario.
Energy Conservation
Energy conservation is a core concept in physics that applies to many types of motion, including inclined planes with or without friction.
The law of conservation of energy states that the total energy in an isolated system remains constant, meaning energy cannot be created or destroyed, only transformed from one form to another.
For a block on an incline, the crucial forms of energy to consider are kinetic energy, potential energy, and sometimes thermal energy due to friction.Here's a quick breakdown:
  • Kinetic Energy (\( KE \)): Energy due to the block's movement, given by \( \frac{1}{2} m v^2 \).
  • Potential Energy (\( PE \)): Energy due to positioning or height, represented as \( mgh \).
  • Thermal Energy: Results from work done against friction.
In solving problems, it's essential to set up your energy equation correctly. For a frictionless plane, you equate kinetic energy at the start to potential energy at the highest point. When friction is present, you account for energy converted into thermal energy as well.
Thermal Energy Increase
The thermal energy increase is a secondary effect of work done by friction during the motion of a block on an inclined plane.
This phenomenon occurs because kinetic friction not only opposes the motion but also transforms some of that mechanical energy into thermal energy.
To calculate the increase in thermal energy, we use:\[ \Delta E_{\text{thermal}} = f_k \cdot d \] This formula indicates that the increase in thermal energy is the product of the frictional force and the distance the block travels against the friction.
In practical terms, the greater the distance and the stronger the friction, the more heat is generated. This concept is crucial for understanding how energy is dissipated in systems involving friction, affecting overall energy conservation as well.

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Most popular questions from this chapter

The pulley has negligible mass, and both it and the inclined plane are frictionless. Block \(A\) has a mass of \(1.0\) \(\mathrm{kg}\), block \(B\) has a mass of \(2.0 \mathrm{~kg}\), and angle \(\theta\) is \(30^{\circ} .\) If the blocks are released from rest with the connecting cord taut, what is their total kinetic energy when block \(B\) has fallen \(25 \mathrm{~cm} ?\)

A volcanic ash flow is moving across horizontal ground when it encounters a \(10^{\circ}\) upslope. The front of the flow then travels 920 \(\mathrm{m}\) up the slope before stopping. Assume that the gases entrapped in the flow lift the flow and thus make the frictional force from the ground negligible; assume also that the mechanical energy of the front of the flow is conserved. What was the initial speed of the front of the flow?

Shows a pendulum of length \(L=1.25 \mathrm{~m}\). Its bob (which effectively has all the mass) has speed \(v_{0}\) when the cord makes an angle \(\theta_{0}=40.0^{\circ}\) with the vertical. (a) What is the speed of the bob when it is in its lowest position if \(v_{0}=8.00 \mathrm{~m} / \mathrm{s} ?\) What is the least value that \(v_{0}\) can have if the pendulum is to swing down and then up (b) to a horizontal position, and (c) to a vertical position with the cord remaining straight? (d) Do the answers to (b) and (c) increase, decrease, or remain the same if \(\theta_{0}\) is increased by a few degrees?

A block of mass \(m=3.20 \mathrm{~kg}\) slides from rest a distance \(d\) down a frictionless incline at angle \(\theta=30.0^{\circ}\) where it runs into a spring of spring constant \(431 \mathrm{~N} / \mathrm{m}\). When the block momentarily stops, it has compressed the spring by \(21.0 \mathrm{~cm} .\) What are (a) distance \(d\) and (b) the distance between the point of the first block-spring contact and the point where the block's speed is greatest?

At a certain factory, \(300 \mathrm{~kg}\) crates are dropped vertically from a packing machine onto a conveyor belt moving at \(1.20 \mathrm{~m} / \mathrm{s}\) (Fig. \(8-62\) ). (A motor maintains the belt's constant speed.) The coefficient of kinetic friction between the belt and each crate is \(0.400\). After a short time, slipping between the belt and the crate ceases, and the crate then moves along with the belt. For the period of time during which the crate is being brought to rest relative to the belt, calculate, for a coordinate system at rest in the factory, (a) the kinetic energy supplied to the crate, (b) the magnitude of the kinetic frictional force acting on the crate, and (c) the energy supplied by the motor. (d) Explain why answers (a) and (c) differ.

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