/*! 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 9 (co A \(3.5 \mathrm{~kg}\) block... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 9

(co A \(3.5 \mathrm{~kg}\) block is pushed along a horizontal floor by a force \(\vec{F}\) of magnitude \(15 \mathrm{~N}\) at an angle \(\theta=40^{\circ}\) with the horizontal (Fig. 6-19). The coefficient of kinetic friction between the block and the floor is \(0.25 .\) Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block's acceleration.

Short Answer

Expert verified
(a) The frictional force is \( f_k = 8.55 \text{ N} \). (b) The block's acceleration is \( a = 1.3 \text{ m/s}^2 \).

Step by step solution

01

Separate Force into Components

Start by resolving the applied force \(\vec{F}\) into its horizontal and vertical components. The horizontal component \( F_{x} \) is given by \( F_x = F \cos(\theta) \), and the vertical component \( F_{y} \) is given by \( F_y = F \sin(\theta) \). For \( F = 15 \text{ N} \) and \( \theta = 40^\circ \), the components are \( F_x = 15 \cos(40^\circ) \) and \( F_y = 15 \sin(40^\circ) \).
02

Calculate Normal Force

The normal force \( N \) is affected by the vertical component of the applied force. Set up the equilibrium condition in the vertical direction: \( N + F_y = mg \), where \( m = 3.5 \text{ kg} \) is the mass and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Solve for \( N \): \( N = mg - F_y \).
03

Calculate the Frictional Force

The frictional force \( f_{ ext{k}} \) is given by the product of the coefficient of kinetic friction \( \mu_k \) and the normal force: \( f_k = \mu_k N \), where \( \mu_k = 0.25 \). Substitute \( N \) from Step 2 into this equation to find \( f_k \).
04

Net Force in the Horizontal Direction

Calculate the net force acting on the block in the horizontal direction. The net force \( F_{ ext{net}} \) is the horizontal component of the applied force minus the frictional force: \( F_{ ext{net}} = F_x - f_k \).
05

Find the Block's Acceleration

Use Newton's second law to find the block's acceleration. The acceleration \( a \) is given by \( a = \frac{F_{ ext{net}}}{m} \), where \( F_{ ext{net}} \) is the result from Step 4 and \( m \) is the block's mass. Solve for \( a \).

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Key Concepts

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

Kinetic Friction
When a block slides across a floor, there's a force that tries to resist its movement. This is called kinetic friction. Kinetic friction depends on:
  • The nature of the surfaces in contact, expressed as the coefficient of kinetic friction \( \mu_k \).
  • The normal force \( N \), which is the force exerted perpendicularly by the surface on the object.
The kinetic frictional force \( f_k \) can be calculated using the formula:
\[ f_k = \mu_k N \]In our exercise, \( \mu_k = 0.25 \) (a typical value for many surfaces).
The normal force, derived in another step, helps us find the friction force.
This resistance plays a crucial role in determining how easily an object moves when pushed.
Normal Force
Normal force is the support force exerted by a surface against an object resting on it. It acts perpendicular to the surface.
  • If you place a book on a table, the table exerts an upward force equal to the weight of the book. This is the normal force.
  • It鈥檚 important in calculations like friction, because more normal force usually means more friction.
For our block:
The applied force affects this normal force.
Because it's applied at an angle, it has a vertical component that pushes the block into the floor.
We can find the normal force \( N \) with:
\[ N = mg - F_y \]where \( m = 3.5 \text{ kg} \) and \( F_y = 15 \sin(40^\circ) \).
Adjusting for the applied force helps us accurately compute the normal force.
Acceleration Calculation
Newton's second law of motion is at the heart of understanding how forces affect an object's motion. It states that:
\[ a = \frac{F_{\text{net}}}{m} \]where \( a \) is acceleration, \( F_{\text{net}} \) is the net force acting on the object, and \( m \) is the object's mass.
To find acceleration:
  • First, resolve any forces into components. This helps in understanding the different effects in horizontal and vertical directions.
  • Calculate the net force \( F_{\text{net}} \) in the direction of interest. Here it鈥檚 horizontal.
  • Use the equation above to find \( a \).
For the block:
Once we know the horizontal force acting minus friction, we easily find the block鈥檚 acceleration.
This tells us how quickly it speeds up or slows down.

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Most popular questions from this chapter

SSM A warehouse worker exerts a constant horizontal force of magnitude \(85 \mathrm{~N}\) on a \(40 \mathrm{~kg}\) box that is initially at rest on the horizontal floor of the warehouse. When the box has moved a distance of \(1.4 \mathrm{~m}\), its speed is \(1.0 \mathrm{~m} / \mathrm{s}\). What is the coefficient of kinetic friction between the box and the fioor?

A cat dozes on a stationary merry-go-round, at a radius of \(5.4\) \(\mathrm{m}\) from the center of the ride. Then the operator turns on the ride and brings it up to its proper turning rate of one complete rotation every \(6.0 \mathrm{~s}\). What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding?

A child places a picnic basket on the outer rim of a merry-goround that has a radius of \(4.6 \mathrm{~m}\) and revolves once every \(30 \mathrm{~s}\). (a) What is the speed of a point on that rim? (b) What is the lowest value of the coefficient of static friction between basket and merrygo-round that allows the basket to stay on the ride?

Suppose the coefficient of static friction between the road and the tires on a car is \(0.60\) and the car has no negative lift. What speed will put the car on the verge of sliding as it rounds a level curve of \(30.5 \mathrm{~m}\) radius?

The terminal speed of a sky diver is \(160 \mathrm{~km} / \mathrm{h}\) in the spreadeagle position and \(310 \mathrm{~km} / \mathrm{h}\) in the nosedive position. Assuming that the diver's drag coefficient \(C\) does not change from one position to the other, find the ratio of the effective cross-sectional area \(A\) in the slower position to that in the faster position.
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