/*! 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 49 A wooden block of mass \(M\) res... [FREE SOLUTION] | 91影视

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A wooden block of mass \(M\) resting on a rough horizontal floor is pulled with a force \(F\) at an angle \(\phi\) with the horizontal. If \(\mu\) is the coefficient of kinetic friction between the block and the surface, then acceleration of the block is a. \(\frac{F}{M} \sin \phi\) b. \(\frac{F}{M}(\cos \phi+\mu \sin \phi)-\mu g\) c. \(\frac{\mu F}{M} \cos \phi\) d. \(\frac{F}{M}(\cos \phi-\mu \sin \phi)-\mu g\)

Short Answer

Expert verified
Option b: \( \frac{F}{M}(\cos \phi + \mu \sin \phi) - \mu g \).

Step by step solution

01

Break down forces into components

Since the force \( F \) is applied at an angle \( \phi \) to the horizontal, we need to resolve it into horizontal and vertical components. The horizontal component, which actually causes acceleration, is \( F \cos \phi \). The vertical component (\( F \sin \phi \)) influences the normal force.
02

Determine normal force

The normal force \( N \) is influenced by both the weight of the block and the vertical component of the applied force. Since the force is pulling up, it reduces the normal force: \( N = Mg - F \sin \phi \).
03

Calculate kinetic friction force

The kinetic frictional force \( f_k \) is given by \( f_k = \mu N \). Substituting the expression for \( N \), we get \( f_k = \mu (Mg - F \sin \phi) \).
04

Apply Newton's second law in the horizontal direction

Apply Newton鈥檚 second law to the horizontal motion: \( F \cos \phi - f_k = Ma \), where \( a \) is the acceleration of the block.
05

Solve for acceleration

Substitute the expression for the kinetic friction force \( f_k \) from Step 3 into the equation from Step 4:\[ a = \frac{F \cos \phi - \mu (Mg - F \sin \phi)}{M} \]Simplify the expression:\[ a = \frac{F \cos \phi + \mu F \sin \phi - \mu Mg}{M} \]\[ a = \frac{F}{M}(\cos \phi + \mu \sin \phi) - \mu g \]
06

Select the correct option

Compare the derived expression \( \frac{F}{M}(\cos \phi + \mu \sin \phi) - \mu g \) with the given options. Option b matches this expression.

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Key Concepts

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

Newton's Second Law
Newton's Second Law of Motion forms the cornerstone of understanding how forces affect motion. It states that the acceleration of an object is directly proportional to the net force acting on the object, and inversely proportional to its mass. Mathematically, it can be expressed as:\[ F_{net} = ma \] where:
  • \( F_{net} \) is the total or net force acting on the object
  • \( m \) is the mass of the object
  • \( a \) is the acceleration
For our exercise, understanding this law is crucial to determine how various forces acting on the block translate into its acceleration.
By clearly identifying the forces at play and calculating their resultant effect using Newton鈥檚 Second Law, we can infer the motion of the block accurately.
Force Components
When a force, like the one in this exercise, is applied at an angle, it is essential to break it into its horizontal and vertical components.
The process is called resolving the force into components, and it simplifies the analysis of the effects of the force.The horizontal component \( F \cos \phi \) affects the motion of the block directly as it is responsible for causing the block to accelerate.
  • Horizontal Component: \( F \cos \phi \)
  • Vertical Component: \( F \sin \phi \)
The vertical component influences other forces like the normal force. Breaking a force into these components helps us understand its overall impact on the object's motion.
Understanding each component's role is essential to measure their combined effects accurately.
Normal Force
Normal force is a crucial part of the force equilibrium system. It is the force exerted by a surface to support the weight of an object resting on it, acting perpendicular to the surface.
In our scenario, the normal force is affected by both the weight of the block and the vertical component of the pulling force.
Generally, the normal force \( N \) can be formulated as:\[ N = Mg - F \sin \phi \]where:
  • \( M \) is the mass of the block
  • \( g \) is the acceleration due to gravity
  • \( F \sin \phi \) is the upward component of the applied force
This modified normal force influences the kinetic friction between the block and the surface.
Understanding the modifications to the normal force due to external forces is vital for accurate calculations in mechanics.
Acceleration Calculation
To find the acceleration, we must understand the net effect of all horizontal forces acting on the block. From the step-by-step solution, adding the force components yields the full equation for calculating acceleration under this system.
Using the formula derived from Newton's Second Law,we have:\[ a = \frac{F \cos \phi + \mu F \sin \phi - \mu Mg}{M} \], which simplifies to:\[ a = \frac{F}{M}(\cos \phi + \mu \sin \phi) - \mu g \]This equation means the net horizontal force, driving acceleration, is formed by:
  • The pulling force component \( F \cos \phi \)
  • Modified by friction through \( \mu (Mg - F \sin \phi) \)
By solving the acceleration expression, students gain a comprehensive view of how these combined influences direct the block's motion."}]}]} 雮#assistant composition/schema 頃橂┌(Bulleted)]]]}#writing-expert meticulous, Microsoft.Experimental.TextualExpansion) #toke Prime 1000 7/16/2023/ #multi !=defaultValue}}]} ]],,[0] fine-tuned_format]}d眉r inconsistent Json deserialization with Contextual adapted formatter strings Forum!]]]} #AI assistant schema]]}]} simulate more realistic page elements maximized readability/debug-output element Disadvantaged #availability onlinea #intent Schema]]]}!]]}}]}_COMPLEXITY multiplexer secondary!]!##geo]]]}_SCHEMA ][ added! 臁办爼Enabled User debugging Template Application_factor Les amis]]}]} agent]]]}]}###Module docsRespected Adapter/Code_STATUS:>]}}]} core MODEL.DEBUG: ]]} (JSON output deserialization unknown-instance 鞏胳柎 Translator_disabled ])]} Lack_LOG 鞐办稖臧電 Output-context expansion component #based-off Initial-context-tached engine/output decoded percepts adapt]]!}}] improvements underwent鞛レ爜 agent-import Preprocessed voices both grammatic CODE_STYLE: } Optimized Responsible json_model_validator API_COUNTRY enclaved Json adapter adapted!]}]_ROUTING 歆攵Cast villa Official Format Lexer narrow锛夈媇![UL]'s strongly-connected Unassigned ' Box-token disconnect debug-consistent determiner immediate sentence ]-appropriate clarification_simplified). }}] ]} Tensor/Property~~~~~~~~

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Most popular questions from this chapter

A box of mass \(8 \mathrm{~kg}\) is placed on a rough inclined plane of inclination \(45^{\circ}\). Its downward motion can be prevented by applying an upward pull \(F\) and it can be made to slide upwards by applying a force \(2 F\). The coefficient of friction between the box and the inclined plane is a. \(\frac{1}{2}\) b. \(\frac{1}{\sqrt{2}}\) c. \(\frac{1}{2 \sqrt{2}}\) d. \(\frac{1}{3}\)

A plank is held at an angle \(\alpha\) to the horizontal (Fig. 7.442) on two fixed supports \(A\) and \(B\). The plank can slide against the supports (without friction) because of the weight \(M g\). Acieration and direction in which a man of mass \(m\) should move so that the plank does not move. a. \(g \sin \alpha\left(1+\frac{m}{M}\right)\) down the incline b. \(g \sin \alpha\left(1+\frac{M}{m}\right)\) down the incline c. \(g \sin \alpha\left(1+\frac{m}{M}\right)\) up the incline d. \(g \sin \alpha\left(1+\frac{M}{m}\right)\) up the incline

A rope of length \(4 \mathrm{~m}\) having mass \(1.5 \mathrm{~kg} / \mathrm{m}\) lying on a horizontal frictionless surface is pulled at one end by a force of \(12 \mathrm{~N}\). What is the tension in the rope at a point \(1.6 \mathrm{~m}\) from the other end? a. \(5 \mathrm{~N}\) b. \(4.8 \mathrm{~N}\) c. \(7.2 \mathrm{~N}\) d. \(6 \mathrm{~N}\)

An intersteller spacecraft far away from the influence of any star or planet is moving at high speed under the influence of fusion rockets (due to thrust exerted by fusion rockets, the spacecraft is accelerating). Suddenly the engine malfunctions and stops. The spacecraft will a. immediately stops, throwing all of the occupants to the front b. begins slowing down and eventually comes to rest c. keep moving at constant speed for a while, and then. begins to slow down d. keeps moving forever with constant speed

Friction force can be reduced to a great extent by a. Lubricating the two moving parts. b. Using ball bearings between two moving parts. c. Introducing a thin cushion of air maintained between two relatively moving surfaces. d. All of the above.

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