/*! 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 8 An empty crate is given an initi... [FREE SOLUTION] | 91Ó°ÊÓ

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An empty crate is given an initial push down a ramp, starting with speed \(v_{0,}\) and reaches the bottom with speed \(v\) and kinetic energy \(K .\) Some books are now placed in the crate, so that the total mass is quadrupled. The coefficient of kinetic friction is constant and air resistance is negligible. Starting again with \(v_{0}\) at the top of the ramp, what are the speed and kinetic energy at the bottom? Explain the reasoning behind your answers.

Short Answer

Expert verified
The speed at the bottom remains \( v \), while kinetic energy becomes \( 4K \).

Step by step solution

01

Understand Initial Conditions

Initially, the crate with its original mass reaches the bottom with a speed of \( v \) and has kinetic energy \( K \). The initial speed at the top of the ramp is \( v_0 \).
02

Analyze the Effect of Mass on Kinetic Energy

When the mass is quadrupled, the kinetic energy, which depends on mass \( (m) \) and speed \( (v) \), will be affected. Recall the formula for kinetic energy: \( K = \frac{1}{2}mv^2 \). With the new mass, \( m_{new} = 4m \).
03

Consider Friction and Energy Conservation

Since the coefficient of kinetic friction and the ramp length remain constant, the work done by friction on the crate is the same in both cases. The gravitational potential energy lost by crates of different masses over the same height also only depends on the mass, meaning the dynamic interactions with gravity stay consistent in proportion.
04

Set Up the Energy Equation for the Increased Mass

The initial kinetic energy plus the potential energy lost minus the work done by friction equals the final kinetic energy. For the empty crate: \( K_{initial} + U - W_{friction} = \frac{1}{2}mv^2 \). For the loaded crate, this becomes \( 4K_{initial} + 4U - 4W_{friction} = \frac{1}{2}(4m)v^2_{new} \).
05

Analyze the Relationship in Terms of Speed

The speed at the bottom for the loaded crate, \( v_{new} \), can be similarly analyzed by considering energy balance at initial condition: both momentum terms keep the system ratios determined by initial and final boundary conditions equal across system variations.
06

Conclusion of the Speed and Kinetic Energy

The speed \( v_{new} \) at the bottom remains \( v \) as friction is consistently scaled by mass, and thus energy conversion balance doesn’t alter speed. Kinetic energy, being proportional to mass for the constant speed ratio, quadruples, becoming \( 4K \).

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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 an object has because of its motion. It is a fundamental concept in mechanics and can be calculated using the formula: \[ K = \frac{1}{2}mv^2 \] where \( m \) is the mass of the object and \( v \) is its velocity.
  • The kinetic energy of a moving object increases with its mass; the heavier it is, the more energy it needs to move.
  • The kinetic energy also increases with the square of its speed, meaning that if the speed doubles, the kinetic energy quadruples.
In our scenario, when the mass of the crate is quadrupled, while starting with the same initial speed \( v_0 \), the kinetic energy at the bottom also quadruples. This is because kinetic energy is directly proportional to mass when the speed is unchanged.
Mass and Inertia
Mass is a measure of the amount of matter in an object, while inertia refers to the tendency of an object to resist changes in its state of motion.
  • When an object's mass increases, its inertia also increases, making it harder to start moving, stop, or change direction.
  • With the original crate's mass, it reaches a certain speed at the bottom of a ramp due to its inertia balancing the forces acting on it.
In our exercise, the mass of the crate is increased by adding books. This doesn’t affect the speed at the bottom of the ramp, as the balance of forces, including friction, remains the same. This is because both gravitational force and friction scale proportionally with mass, counterbalancing any changes in inertia.
Friction
Friction is the force that opposes motion between two surfaces in contact. It plays a crucial role in mechanics and can affect how fast or slow an object moves.
  • In this exercise, kinetic friction is considered, which is constant as given by a coefficient in the scenario.
  • The frictional force is calculated as \( f_k = \mu_k \cdot n \), where \( \mu_k \) is the coefficient of kinetic friction and \( n \) is the normal force.
Since the mass of the crate has changed, the normal force has also increased, therefore the friction increases too. Yet, because both the gravitational force that pulls the crate down and the friction that opposes this motion increase proportionally, the speed at the bottom remains unchanged.
Energy Conservation
Energy conservation is a principle in physics stating that the total energy of an isolated system remains constant. It is crucial in analyzing systems where energy can transform from one form to another, such as potential energy to kinetic energy in this scenario.
  • Initially, the crate has potential energy due to its height on the ramp, which gets converted to kinetic energy as it descends.
  • The presence of friction means some of this energy is also spent overcoming friction.
In our problem, even though the crate’s mass increases, the principle of energy conservation means that the energy transformations happen proportionally to the mass. After accounting for energy spent on friction (which also scales with mass), the principle ensures that such energy loss doesn’t alter the final speed. Finally, the quadrupled mass results in quadrupled kinetic energy, showcasing how energy conservation maintains consistent outcomes despite variations in mass in an otherwise unchanged system.

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Most popular questions from this chapter

CP A 0.300 -kg potato is tied to a string with length \(2.50 \mathrm{m},\) and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. (a) What is the speed of the potato at the lowest point of its motion? (b) What is the tension in the string at this point?

A 2.00 -kg block is pushed against a spring with negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m},\) compressing it 0.220 \(\mathrm{m}\) . When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37.0" (Fig. \(\mathrm{P7} .42\) ). (a) What is the speed of the block as it slides along the horizontal surface after having left the spring? (b) How far does the block travel up the incline before starting to slide back down?

CALC In an experiment, one of the forces exerted on a proton is \(\vec{\boldsymbol{F}}=-\alpha x^{2} \hat{\boldsymbol{r}},\) where \(\alpha=12 \mathrm{N} / \mathrm{m}^{2}\) . (a) How much work does \(\vec{\boldsymbol{F}}\) do when the proton moves along the straight-line path from the point \((0.10 \mathrm{m}, 0)\) to the point \((0.10 \mathrm{m}, 0.40 \mathrm{m}) ?(\mathrm{b})\) Along the straight-line path from the point \((0.10 \mathrm{m}, 0)\) to the point \((0.30 \mathrm{m}, 0) ?\) (c) Along the straight-line path from the point \((0.30 \mathrm{m}, 0)\) to the point \((0.10 \mathrm{m}, 0) ?(\mathrm{d})\) Is the force \(\vec{\boldsymbol{F}}\) conservative? Explain. If \(\vec{\boldsymbol{F}}\) is conservative, what is the potential-energy function for it? Let \(U=0\) when \(x=0 .\)

A wooden block with mass 1.50 \(\mathrm{kg}\) is placed against a compressed spring at the bottom of an incline of slope \(30.0^{\circ}\) (point \(A )\) . When the spring is released, it projects the block up the incline. At point \(B\) , a distance of 6.00 \(\mathrm{m}\) up the incline from \(A\) . the block is moving up the incline at 7.00 \(\mathrm{m} / \mathrm{s}\) and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_{\mathrm{k}}=0.50 .\) The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

A 0.60 -kg book slides on a horizonal table. The kinetic friction force on the book has magnitude 1.2 \(\mathrm{N}\) . (a) How much work is done on the book by friction during a displacement of 3.0 \(\mathrm{m}\) to the left? (b) The book now slides 3.0 \(\mathrm{m}\) to the right, returning to its starting point. During this second 3.0 -m displacement, how much work is done on the book by friction? (c) What is the total work done on the book by friction during the complete round trip? (d) On the basis of your answer to part \((c),\) would you say that the friction force is conservative or nonconservative?Explain.

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