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Chapter 5: Problem 30

Some sliding rocks approach the base of a hill with a speed of \(12 \mathrm{~m} / \mathrm{s} .\) The hill rises at \(36^{\circ}\) above the horizontal and has coefficients of kinetic friction and static friction of 0.45 and \(0.65,\) respectively, with these rocks. (a) Find the acceleration of the rocks as they slide up the hill. (b) Once a rock reaches its highest point, will it stay there or slide down the hill? If it stays, show why. If it slides, find its acceleration on the way down.

Short Answer

Expert verified
The solution to this exercise will be dependent on specific calculations, and may vary depending on given mass of the rocks (if any). However, the steps presented will guide through the process of solving this problem.

Step by step solution

01

Identify Given Values

The initial speed of sliding rocks is 12 m/s. The angle of hill inclination above the horizontal is \(36^{\circ}\). The coefficients of kinetic and static friction are 0.45 and 0.65, respectively.
02

Calculate the Acceleration Up the Hill

To find the acceleration of the rocks as they slide up the hill, start with Newton's second law, which states that force equals mass times acceleration (F=ma). The frictional force is \(f_k = \mu_k \* mg \* cos(36^{\circ})\) and the force from gravity is \(F_g = mg \* sin(36^{\circ})\). The net force acting on the rock is the difference between gravity and friction: \(F_net = F_g - f_k = ma\). Solve for acceleration (a) to get the acceleration of the rocks up the hill.
03

Find Behavior at the Highest Point

When the rock reaches its highest point, its kinetic energy is zero but it still has potential energy due to its height. The rock will either stay at its highest point or will begin to slide down the hill. This is determined by comparing the gravitational force pulling it down the slope \(F_g = mg \* sin(36^{\circ})\) and the static friction force resisting motion \(f_s = \mu_s \* mg \* cos(36^{\circ})\). If the static frictional force is larger than or equal to the gravitational force, the rock will stay. If the gravitational force is larger, the rock will slide.
04

Calculate the Acceleration Down the Hill

If the rock begins to slide down the hill, calculate the acceleration down the hill using Newton's second law. The net force acting on the rock is the difference between the gravitational force and the frictional force: \(F_net = F_g - f_k = ma\). Solve for 'a' to get the acceleration of the rock down the hill.

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Key Concepts

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

Kinetic Friction
When objects are in motion, a force that tries to resist their relative movement across a surface is at work. This force is called kinetic friction, and it's an essential part of Newton's Laws of Motion. In the context of sliding rocks on an inclined plane, kinetic friction plays a critical role.
Kinetic friction depends on two factors: the nature of the surfaces in contact and how hard they are pressed together. The force of kinetic friction can be calculated using the formula:
  • \(f_k = \mu_k \times N\)
  • \(N\) is the normal force, or the perpendicular force exerted by a surface against an object resting on it.
  • \(\mu_k\) is the coefficient of kinetic friction, which quantifies how much friction force exists between the surfaces.
For sliding rocks on a hill inclined at \(36^{\circ}\), the **normal force** is \(mg \times \cos(36^{\circ})\), where \(mg\) is the gravitational force. Thus, the kinetic friction is calculated as \(f_k = \mu_k \times mg \times \cos(36^{\circ})\). This force opposes the downhill slide, affecting the rocks' acceleration.
Static Friction
Before an object starts moving on a surface, it is often held in place by a type of friction called static friction. This frictional force is usually stronger than kinetic friction and works to keep stationary objects from beginning to move.
When we consider rocks at the maximum height on an inclined plane, static friction comes into play to determine if they'll stay put or start sliding down.
The formula for static friction is:
  • \(f_s = \mu_s \times N\)
  • \(\mu_s\) is the coefficient of static friction, which is typically higher than \(\mu_k\).
Comparing the gravitational pull down the slope \(mg \times \sin(36^{\circ})\) to the static friction force \(f_s = \mu_s \times mg \times \cos(36^{\circ})\) helps us decide whether the rocks will remain at rest. If static friction is greater or equal to the gravitational force, the rocks will stay put. Otherwise, they will start sliding, and kinetic friction will again become relevant.
Inclined Plane
An inclined plane is a flat surface tilted at an angle relative to the horizontal. It's a fundamental mechanical scenario for studying forces and motion. When objects like rocks slide on such surfaces, gravity components, friction, and normal forces intertwine in intriguing ways.
Studying an inclined plane can shed light on how objects accelerate or decelerate, thanks to the interplay of these forces. The crucial angle of the plane influences how strong each force is:
  • The **gravitational force** down the slope is given by \(mg \times \sin(\theta)\), where \(\theta\) is the angle of incline.
  • The **normal force** is \(mg \times \cos(\theta)\).
Analyzing the forces acting on an object on an inclined plane allows us to predict its motion. For example, determining whether a rock will stop, rest, or slide back down after reaching the highest point occurs through balancing these forces. Gravity works to pull the object along the slope, while friction seeks to resist this motion.

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Most popular questions from this chapter

Two blocks, with masses \(4.00 \mathrm{~kg}\) and \(8.00 \mathrm{~kg},\) are connected by a string and slide down a \(30.0^{\circ}\) inclined plane (Fig. \(\mathbf{P 5 . 9 6}\) ). The coefficient of kinetic friction between the \(4.00 \mathrm{~kg}\) block and the plane is \(0.25 ;\) that between the \(8.00 \mathrm{~kg}\) block and the plane is \(0.35 .\) Calculate (a) the acceleration of each block and (b) the tension in the string. (c) What happens if the positions of the blocks are reversed, so that the \(4.00 \mathrm{~kg}\) block is uphill from the \(8.00 \mathrm{~kg}\) block?

A \(3.00 \mathrm{~kg}\) box that is several hundred meters above the earth's surface is suspended from the end of a short vertical rope of negligible mass. A time-dependent upward force is applied to the upper end of the rope and results in a tension in the rope of \(T(t)=(36.0 \mathrm{~N} / \mathrm{s}) t\) The box is at rest at \(t=0 .\) The only forces on the box are the tension in the rope and gravity. (a) What is the velocity of the box at (i) \(t=1.00 \mathrm{~s}\) and (ii) \(t=3.00 \mathrm{~s} ?\) (b) What is the maximum distance that the box descends below its initial position? (c) At what value of \(t\) does the box return to its initial position?

Friction and Climbing Shoes. Shoes made for the sports of bouldering and rock climbing are designed to provide a great deal of friction between the foot and the surface of the ground. Such shoes on smooth rock might have a coefficient of static friction of 1.2 and a coefficient of kinetic friction of 0.90 . For a person wearing these shoes, what's the maximum angle (with respect to the horizontal) of a smooth rock that can be walked on without slipping? (a) \(42^{\circ} ;\) (b) \(50^{\circ} ;\) (c) \(64^{\circ} ;\) (d) larger than \(90^{\circ}\).

Two crates connected by a rope lie on a horizontal surface (Fig. E5.37). Crate \(A\) has mass \(m_{A}\), and crate \(B\) has mass \(m_{B}\). The coefficient of kinetic friction between each crate and the surface is \(\mu_{\mathrm{k}} .\) The crates are pulled to the right at constant velocity by a horizontal force \(\overrightarrow{\boldsymbol{F}}\). Draw one or more free-body diagrams to calculate the following in terms of \(m_{A}, m_{B},\) and \(\mu_{\mathrm{k}}:\) (a) the magnitude of \(\overrightarrow{\boldsymbol{F}}\) and \((\mathrm{b})\) the tension in the rope connecting the blocks.

A rocket of initial mass \(125 \mathrm{~kg}\) (including all the contents) has an engine that produces a constant vertical force (the thrust) of \(1720 \mathrm{~N}\). Inside this rocket, a \(15.5 \mathrm{~N}\) electric power supply rests on the floor. (a) Find the initial acceleration of the rocket. (b) When the rocket initially accelerates, how hard does the floor push on the power supply? (Hint: Start with a free-body diagram for the power supply.)
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