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Chapter 7: Problem 48

Up and Down the Hill. \(\wedge 28\) -kg rock approaches the foot of a hill with a speed of 15 \(\mathrm{m} / \mathrm{s}\) . This hill slopes upward at a constant angle of \(40.0^{\circ}\) above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20 , respectively. (a) Use energy conservation to find the maximum height above the foot of the hill reached by the rock. (b) Will the rock remain at rest at its highest point, or will it slide back down the hill? (c) If the rock does slide back down, find its speed when it returns to the bottom of the hill.

Short Answer

Expert verified
The rock reaches approximately 6.66 meters. It does not slide back due to static friction.

Step by step solution

01

Understanding the Problem

We need to determine how high the rock climbs the hill by applying energy conservation principles. This involves calculating the energy conversion from kinetic to potential energy and accounting for energy loss due to friction.
02

Initial and Final Energies

The initial energy of the system is the kinetic energy of the rock at the bottom of the hill. This is given by \( KE_i = \frac{1}{2}mv^2 \), where \( m = 28 \text{ kg} \) and \( v = 15 \text{ m/s} \). The potential energy at the height \( h \) is \( PE_f = mgh \). The frictional force will do work on the rock as it climbs, which is given by \( W_f = \mu_kmg\cos(\theta)d \), where \( \mu_k = 0.20 \). We set \( \theta = 40^\circ \).
03

Equation for Energy Conversion

Apply energy conservation: the initial kinetic energy minus the work done by friction equals the potential energy at height \( h \). This can be expressed as:\[ \frac{1}{2}mv^2 - \mu_kmg\cos(\theta)d = mgh \]We need to solve for \( h \), where \( d = h/\sin(\theta) \) is the distance along the slope.
04

Simplify and Solve for Height

Substituting \( d = h/\sin(\theta) \) in the equation:\[ \frac{1}{2}mv^2 - \mu_kmg\cos(\theta)\frac{h}{\sin(\theta)} = mgh \]Solving for \( h \):\[ h = \frac{\frac{1}{2}v^2}{g(\sin(\theta) + \mu_k\cos(\theta))} \]Plug in the values \( v = 15 \text{ m/s}, \theta = 40^\circ, \mu_k = 0.20, \text{ and } g = 9.8 \text{ m/s}^2 \) to find \( h \).
05

Calculate Maximum Height

Calculating using the equation from Step 4:\[ h = \frac{\frac{1}{2} \times 15^2}{9.8 \times (\sin(40^\circ) + 0.20 \times \cos(40^\circ))} \]This simplifies to:\[ h \approx 6.66 \text{ meters} \]
06

Determine If the Rock Slides Back

To determine if it slides back, compare static friction to gravitational pull down the slope at the top. The force due to gravity parallel to the slope is \( mg\sin(\theta) \) and the maximum static friction is \( \mu_s mg\cos(\theta) \), where \( \mu_s = 0.75 \). Since \( \mu_s mg\cos(\theta) > mg\sin(\theta) \), the static friction is sufficient to keep the rock at rest.
07

Conclusion Regarding the Rock's Motion

Due to sufficient static friction, the rock remains at rest at its highest point and does not slide back down the hill. Hence, part (c) of the problem does not apply.

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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 of an object due to its motion. When the rock is at the bottom of the hill, it possesses a certain amount of kinetic energy because of its speed. This energy is calculated using the formula: \[ KE_i = \frac{1}{2}mv^2 \]where \( m \) is the mass of the rock (28 kg), and \( v \) is the speed (15 m/s). When substituted, this gives the initial kinetic energy, which is the starting energy that will eventually convert into potential energy as the rock climbs the hill.
  • Kinetic energy depends on both the mass and the velocity of the object.
  • More speed or mass leads to higher kinetic energy.
The importance of kinetic energy in this problem lies in understanding how this initial energy helps the rock move upwards, fighting against gravitational forces. And this is why kinetic energy is a key piece in the wider puzzle of energy conservation.
potential energy
Potential energy is the energy stored in an object because of its position or configuration. As the rock ascends the hill, its kinetic energy is transformed into gravitational potential energy. This transformation occurs because the rock is moving against gravity. Potential energy at a height \( h \) is calculated by:\[ PE_f = mgh \]where \( m \) is the mass of the rock (28 kg), \( g \) is the acceleration due to gravity (9.8 m/s虏), and \( h \) is the height reached.
  • Potential energy increases with height.
  • A strong gravitational field (like Earth's) increases potential energy.
In the scenario of the rock, potential energy tells us how high up it can climb before gravity stops it. As the rock moves higher, kinetic energy falls while potential energy rises, ensuring the total mechanical energy of the system remains constant (ignoring losses like friction).
friction
Friction is a resistive force that acts against the motion of an object. In the scenario of the rock climbing a hill, we consider two types of friction: static and kinetic. Kinetic friction comes into play as the rock is moving up the hill, opposing its motion and reducing kinetic energy.The force of kinetic friction is given by:\[ W_f = \mu_kmg\cos(\theta)d \]where \( \mu_k \) is the coefficient of kinetic friction (0.20), \( \theta \) is the slope angle (40掳), and \( d \) is the distance traveled on the slope.
  • Friction transforms kinetic energy into heat, reducing the efficient motion upwards.
  • Static friction, however, will keep the rock stationary at the height once it stops.
Understanding friction helps us calculate how much energy is lost during the rock's ascent. Additionally, friction ensures that the rock stays at rest at the top rather than sliding back, providing a fascinating look at how resistive forces impact movement and stability.

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Most popular questions from this chapter

CALE A 3.00 -kg fish is attached to the lower end of a vertical spring that has negligible mass and force constant 900 \(\mathrm{N} / \mathrm{m}\) . The spring initially is neither stretched nor compressed. The fish is released from rest. (a) What is its speed after it has descended 0.0500 \(\mathrm{m}\) from its initial position? (b) What is the maximum speed of the tish as it descends?

A wooden rod of negligible mass and length 80.0 \(\mathrm{cm}\) is pivoted about a horizontal axis through its center. A white rat with mass 0.500 kg clings to one end of the stick, and a mouse with mass 0.200 kg clings to the other end. The system is released from rest with the rod horizontal. If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position?

CALC The potential energy of two atoms in a diatomic molecule is approximated by \(U(r)=a / r^{12}-b / r^{6},\) where \(r\) is the spacing between atoms and \(a\) and \(b\) are positive constants. (a) Find the force \(F(r)\) on one atom as a function of \(r .\) Draw two graphs. one of \(U(r)\) versus \(r\) and one of \(F(r)\) versus \(r\) . \(b\) ) Find the equilibrium distance between the two atoms. Is this equilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it - that is, to separate the two atoms to an infinite distance apart? This is called the dissociation energy of the molecule. (d) For the molecule CO, the equilibrium distance between the carbon and oxygen atoms is \(1.13 \times 10^{-10} \mathrm{m}\) and the dissociation energy is \(1.54 \times 10^{-18} \mathrm{J}\) per molecule. Find the values of the constants \(a\) and \(b .\)

A 2.50 -kg mass is pushed against a horizontal spring of force constant 25.0 \(\mathrm{N} / \mathrm{cm}\) on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring in any way. When the spring has been compressed enough to store 11.5 \(\mathrm{J}\) of potential energy in it, the mass is suddenly released from rest. (a) Find the greatest speed the mass reaches. When does this occur? (b) What is the greatest acceleration of the mass, and when does it occur?

The Great Sandini is a 60 -kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1100 \(\mathrm{N} / \mathrm{m} / \mathrm{m}\) that he will compress with a force of 4400 \(\mathrm{N}\) .The inside of the gun barrel is coated with Teflon, so the average friction force will be only 40 \(\mathrm{N}\) during the 4.0 \(\mathrm{m}\) he moves in the barrel. At what speed will he emerge from the end of the barrel, 2.5 \(\mathrm{m}\) above his initial rest position?
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