Ch 06: Dynamics I: Motion Along a Line

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 06: Dynamics I: Motion Along a Line

Problem 55

Chapter 6, Problem 55

A large box of mass M is moving on a horizontal surface at speed v₀. A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μ_s and μ_k, respectively. Find an expression for the shortest distance d_min in which the large box can stop without the small box slipping.

Verified step by step guidance

Identify the forces acting on the small box: The small box experiences a gravitational force downward (mg), a normal force upward (N), and a frictional force (f) due to the large box. The frictional force is what prevents the small box from slipping.

Determine the maximum static friction force: The maximum static friction force is given by

f_{\(\text{max}\)} = \(\mu\)_s N

. Since the normal force is equal to the weight of the small box,

N = mg

, we have

f_{\(\text{max}\)} = \(\mu\)_s mg

Relate the frictional force to the acceleration: For the small box to not slip, the frictional force must provide the necessary force to accelerate the small box at the same rate as the large box. Using Newton's second law,

f = ma

, where

a

is the acceleration of the large box. Thus,

a \(\leq\) \(\mu\)_s g

Relate the acceleration to the stopping distance: The large box decelerates uniformly to stop. Using the kinematic equation

v_f^2 = v_0^2 + 2ad

, where

v_f = 0

(final velocity),

v_0

is the initial velocity,

a

is the deceleration, and

d

is the stopping distance, we get

0 = v_0^2 - 2ad

. Solving for

d

, we find

d = \(\frac{v_0^2}{2a}\)

Substitute the maximum allowable acceleration: Replace

a

with

\(\mu\)_s g

to ensure the small box does not slip. The shortest stopping distance is then

d_{\(\text{min}\)} = \(\frac{v_0^2}{2\mu_s g}\)

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Friction

Friction is the force that opposes the relative motion of two surfaces in contact. It is characterized by two coefficients: static friction (μₛ), which prevents motion until a certain threshold is reached, and kinetic friction (μₖ), which acts when the surfaces are sliding against each other. Understanding these coefficients is crucial for determining the conditions under which the small box will remain stationary on the large box as it decelerates.

Newton's Second Law

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma). This principle is essential for analyzing the forces acting on both the large box and the small box, allowing us to calculate the deceleration required to stop the large box without causing the small box to slip.

Relative Motion

Relative motion refers to the motion of an object as observed from another moving object. In this scenario, it is important to analyze the motion of the small box relative to the large box. By understanding how the deceleration of the large box affects the small box, we can derive the minimum stopping distance that prevents slipping, ensuring that the small box remains stationary on top of the large box.

Recommended video:

Guided course

04:27

Intro to Relative Motion (Relative Velocity)

Related Practice

Textbook Question

The 2.0 kg wood box in FIGURE P6.58 slides down a vertical wood wall while you push on it at a 45° angle. What magnitude of force should you apply to cause the box to slide down at a constant speed?

views

Textbook Question

Astronauts in space 'weigh' themselves by oscillating on a spring. Suppose the position of an oscillating 75 kg astronaut is given by $x = (0.30 m) \sin ((π rad/s) \times t)$ , where t is in s. What force does the spring exert on the astronaut at (a) t = 1.0 s and (b) 1.5 s? Note that the angle of the sine function is in radians.

views

Textbook Question

A 1.0 kg wood block is pressed against a vertical wood wall by the 12 N force shown in FIGURE P6.57. If the block is initially at rest, will it move upward, move downward, or stay at rest?

Textbook Question

A baggage handler drops your 10 kg suitcase onto a conveyor belt running at 2.0 m/s. The materials are such that μ_s = 0.50 and μ_k = 0.30. How far is your suitcase dragged before it is riding smoothly on the belt?

Textbook Question

A 5.0 kg wooden sled is launched up a 25° snow-covered slope with an initial speed of 10 m/s. What vertical height does the sled reach above its starting point?

views

Textbook Question

Sam, whose mass is 75 kg, takes off across level snow on his jet-powered skis. The skis have a thrust of 200 N and a coefficient of kinetic friction on snow of 0.10. Unfortunately, the skis run out of fuel after only 10 s. How far has Sam traveled when he finally coasts to a stop?