/*! 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 36 \(\mathrm{}\) A \(25.0 \mathrm{~... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\mathrm{}\) A \(25.0 \mathrm{~kg}\) box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is \(0.25,\) and the coefficient of static friction is \(0.35 .(\) a) As \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid \(5.0 \mathrm{~m}\) along the loading ramp?

Short Answer

Expert verified
The minimum angle at which the box starts to slip is \(19.3^{\circ}\), the acceleration of the box when it starts moving is \(0.73 m/s^2\), and the speed of the box after it has slid 5.0m along the ramp is \(3.42 m/s\).

Step by step solution

01

Find the minimum angle at which the box starts to slip

The forces at equilibrium are the box's weight acting down the slope and the frictional force acting up the slope. The frictional force at the point of slipping is the maximum static friction. Using the equation for static friction \( f_{smax} = \mu_{s} \cdot N \), with the normal force \( N = mg \cos (\alpha) \) and \( f_{smax} = mg \sin(\alpha) \). These forces balance at equilibrium, hence \( f_{smax} = mg \sin(\alpha) = \mu_{s} \cdot N = \mu_{s} \cdot mg \cos (\alpha) \). Solving this for \( \alpha \) gives \( \alpha = \arctan (\mu_{s}) \). Substituting \( \mu_{s} = 0.35 \) results in \( \alpha \approx 19.3^{\circ} \).
02

Find the acceleration once the box has begun to move

Once the box starts to slip, kinetic friction \( f_k = \mu_k \cdot N \) takes over. Therefore, the net force acting on the box is the difference between the gravitational component going down the slope and the kinetic friction. Using Newton’s Second Law, \( f_{net} = ma \), where \( f_{net} = mg \sin(\alpha) - f_k \), substituting \( f_k = \mu_k \cdot N = \mu_k \cdot mg \cos (\alpha) \) into this equation and solving for the acceleration \( a \) yields \( a = g(\sin(\alpha)- \mu_k \cos (\alpha)) \). Substituting \( \mu_k = 0.25 \), \( g = 9.8 \frac{m}{s^2} \) and \( \alpha = 19.3^{\circ} \) will yield \( a \approx 0.73 \frac{m}{s^2} \) .
03

Find the speed of the box after it has slid 5.0m

The box's speed as it slides down the ramp can be determined by using one of the equations of motion: \( v = u + at \). Since the box starts from rest, the initial speed \( u = 0 \frac{m}{s} \). Hence, the equation reduces to \( v = at \). However, we don't have the time but we know the displacemen (s). We will use another kinematic equation to solve for v: \( v^2 = u^2 + 2as \). Substituting \( u = 0 \frac{m}{s} \), \( a = 0.73 \frac{m}{s^2} \) and \( s = 5.0 m \) will give \( v \approx 3.42 \frac{m}{s} \) .

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinetic Friction
Kinetic friction comes into play when an object is already in motion. It acts opposite to the direction of the movement, making it a crucial factor in slowing down or resisting a moving object's motion on a surface.
In the context of our exercise, once the box of textbooks starts moving on the inclined plane, kinetic friction becomes the dominant force acting uphill against the box.
  • The formula for kinetic friction is given by: \[f_k = \mu_k \cdot N\]where \(\mu_k\) is the coefficient of kinetic friction, and \(N\) is the normal force.
  • In this case, we use the kinetic friction coefficient of \(0.25\) to calculate how much force is resisting the downhill motion once slipping begins.
Understanding kinetic friction allows us to calculate the box’s acceleration once it starts moving, an essential step in solving this type of physics problem.
Static Friction
Before an object starts moving, static friction is what keeps it in place. Static friction is generally stronger than kinetic friction and needs to be overcome for motion to initiate. In our exercise, static friction is pivotal to determining at what angle the box starts sliding down the ramp.
  • The maximum static friction can be calculated using:\[f_{smax} = \mu_s \cdot N\]where \(\mu_s\) is the coefficient of static friction, and \(N\) is the normal force (which can be calculated from the object's weight component perpendicular to the slope).
  • Solving for the angle \(\alpha\) where static friction equals the gravitational component down the slope provides the critical angle for slippage. As seen, \(\mu_s = 0.35\) yields \(\alpha \approx 19.3^\circ\).
Static friction plays a crucial role in preventing motion, and calculating it helps reveal when motion begins.
Newton's Second Law
Newton's Second Law is fundamental in describing the relationship between an object's mass, its acceleration, and the applied force. The law is succinctly captured by the equation:
\[F_{net} = ma\]Where:
  • \(F_{net}\) is the net force acting on the object,
  • \(m\) is the object's mass,
  • \(a\) is the acceleration.
In our scenario, Newton's Second Law helps us understand how forces combine to set the box in motion once it starts slipping. The gravitational force pulling the box down the slope is reduced by kinetic friction, and this resulting net force allows us to compute the box's acceleration when substituting the known values:
\[a = g(\sin(\alpha) - \mu_k\cos(\alpha))\]
This equation highlights Newton's insight into how unbalanced forces result in acceleration.
Kinematics
Kinematics allows us to describe motion through key equations, irrespective of the forces causing it. Once the box on the ramp begins to accelerate, kinematics equations provide crucial insights. These equations link initial velocity, final velocity, acceleration, and displacement to predict motion.
One vital equation from kinematics is:\[v^2 = u^2 + 2as\]where:
  • \(v\) is final velocity,
  • \(u\) is initial velocity,
  • \(a\) is acceleration,
  • \(s\) is displacement.
In our case, with \(u = 0\) and knowing the distance \(s = 5.0 m\), this equation gives us the speed of the box after it has traveled this distance. By plugging in the calculated acceleration, we find that the box reaches a speed of approximately \(3.42 \frac{m}{s}\). Understanding kinematics is crucial for predicting how objects move under certain conditions.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, other than a right angle, to a horizontal surface. It's a classic tool in physics problems because it introduces a simple yet effective way to study forces acting on an object being pulled down by gravity on a slope.
For this exercise, the inclined plane implies:
  • Different force components: gravitational force resolves into two components—parallel and perpendicular to the plane.
  • The normal force acts perpendicular to the plane, affecting how both static and kinetic friction are calculated.
  • The angle of inclination directly affects whether the object remains static or begins to move and, once it moves, impacts its acceleration down the ramp.
An inclined plane provides a practical scenario to explore these physics concepts and demonstrates how angles can drastically change the motion and force dynamics involved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A light rope is attached to a block with mass \(4.00 \mathrm{~kg}\) that rests on a friction less, horizontal surface. The horizontal rope passes over a friction less, massless pulley, and a block with mass \(m\) is suspended from the other end. When the blocks are released, the tension in the rope is \(15.0 \mathrm{~N}\). (a) Draw two free-body diagrams: one for each block. (b) What is the acceleration of either block? (c) Find \(m\). (d) How does the tension compare to the weight of the hanging block?

\(\mathrm{A} 45.0 \mathrm{~kg}\) crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds \(313 \mathrm{~N}\). Then you must reduce your push to \(208 \mathrm{~N}\) to keep it moving at a steady \(25.0 \mathrm{~cm} / \mathrm{s}\). (a) What are the coefficients of static and kinetic friction between the crate and the floor? (b) What push must you exert to give it an acceleration of \(1.10 \mathrm{~m} / \mathrm{s}^{2} ?\) (c) Suppose you were performing the same experiment on the moon, where the acceleration due to gravity is \(1.62 \mathrm{~m} / \mathrm{s}^{2}\). (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?

BIO Force During a Jump. When jumping straight up from a crouched position, an average person can reach a maximum height of about \(60 \mathrm{~cm} .\) During the jump, the person's body from the knees up typically rises a distance of around \(50 \mathrm{~cm} .\) To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. (a) With what initial speed does the person leave the ground to reach a height of \(60 \mathrm{~cm} ?\) (b) Draw a free-body diagram of the person during the jump. (c) In terms of this jumper's weight \(w\), what force does the ground exert on him or her during the jump?

A horizontal wire holds a solid uniform ball of mass \(m\) in place on a tilted ramp that rises \(35.0^{\circ}\) above the horizontal. The surface of this ramp is perfectly smooth, and the wire is directed away from the center of the ball (Fig. P5.64). (a) Draw a free-body diagram of the ball. (b) How hard does the surface of the ramp push on the ball? (c) What is the tension in the wire?

You place a book of mass \(5.00 \mathrm{~kg}\) against a vertical wall. You apply a constant force \(\overrightarrow{\boldsymbol{F}}\) to the book, where \(F=96.0 \mathrm{~N}\) and the force is at an angle of \(60.0^{\circ}\) above the horizontal (Fig. \(\left.\mathbf{P} 5.75\right)\). The coefficient of kinetic friction between the book and the wall is \(0.300 .\) If the book is initially at rest, what is its speed after it has traveled \(0.400 \mathrm{~m}\) up the wall?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.