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

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Chapter 6: Problem 9

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
The frictional force is approximately 7.6 N, and the block's acceleration is approximately 0.92 m/s虏.

Step by step solution

01

Determine the Normal Force

First, calculate the normal force acting on the block. We know the block is subjected to its weight and the vertical component of force \(F\). The weight of the block is \(mg = 3.5 \times 9.8 = 34.3\, \text{N}\). The vertical component of the force \(F\) is given by \(F_Y = F \sin \theta = 15 \times \sin 40^\circ\). The normal force \(N\) is then: \[ N = mg - F \sin \theta = 34.3 - 15 \times \sin 40^\circ. \]
02

Calculate the Frictional Force

Using the normal force calculated earlier, compute the frictional force using the formula: \[ f_k = \mu_k N \] where \(\mu_k = 0.25\) is the coefficient of kinetic friction. Substitute the value of \(N\) from Step 1 to find \(f_k\).
03

Calculate the Net Horizontal Force

Determine the horizontal component of \(\vec{F}\) which is responsible for accelerating the block. It is given by \[ F_X = F \cos \theta = 15 \times \cos 40^\circ\].The net force \(F_{ ext{net}}\) acting on the block is then:\[ F_{\text{net}} = F_X - f_k \] where \(f_k\) is the frictional force from Step 2.
04

Compute the Block's Acceleration

Finally, calculate the acceleration \(a\) of the block using Newton's second law: \[ F_{\text{net}} = ma \]where \(m = 3.5\, \text{kg}\). Solve the equation for \(a\) to find:\[ a = \frac{F_{\text{net}}}{m}. \]

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Key Concepts

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

Frictional Force
When dealing with objects moving along surfaces, frictional force plays a crucial role. Frictional force opposes the motion or attempted motion of an object in contact with a surface.
It always acts in the opposite direction to the applied force. In this exercise, the force you need to consider is the kinetic friction since the block is sliding over the floor.

The frictional force, or more specifically, the kinetic frictional force, is calculated using the formula:
  • \( f_k = \mu_k N \)
Where:
  • \( f_k \) is the kinetic frictional force.
  • \( \mu_k \) is the coefficient of kinetic friction; in this case, it's given as 0.25.
  • \( N \) is the normal force, which is the perpendicular contact force exerted by a surface on an object.
Normal Force
The normal force is fundamental in understanding motion on surfaces. It represents the perpendicular force that a surface exerts on an object resting on it.
In the exercise scenario, this is the force the floor exerts upward against the block.

To calculate the normal force, consider both the weight of the block and any vertical component of any applied forces:
  • The weight of the block is calculated as \( mg \), where \( m \) is the mass (3.5 kg) and \( g \) is the acceleration due to gravity, approximately 9.8 m/s虏.
  • The vertical component of the applied push force is \( F \sin \theta \). This component works in opposition to the block's weight.
  • Thus, the normal force \( N \) is computed as: \( N = mg - F \sin \theta \).
Remember, understanding the normal force is key to determining the frictional force since it directly influences it.
Kinetic Friction
Kinetic friction is the type of friction that acts between moving surfaces. When an object, like our block, slides over a surface, kinetic friction comes into play.
It depends on the characteristics of the surfaces in contact, such as roughness and material.

Kinetic friction is calculated using the normal force and the coefficient of kinetic friction as follows:
  • \( f_k = \mu_k N \)
For this block, we know:
  • \( \mu_k \) is already provided as 0.25.
  • The normal force \( N \) has been previously calculated.
By substituting these values into the equation, you can find the force opposing the block's motion. Kinetic friction is crucial as it defines how difficult it is to keep the object moving.
Acceleration Calculation
Finally, to find the block's acceleration, you use Newton's Second Law of Motion. This law connects the net force acting on an object and its acceleration through:
  • \( F_{\text{net}} = ma \)
To determine the block's acceleration:
  • First, calculate the horizontal component of the applied force: \( F_X = F \cos \theta \).
  • Subtract the kinetic frictional force, \( f_k \), from this horizontal component to find the net force: \( F_\text{net} = F_X - f_k \).
  • Then, rearrange the equation to solve for acceleration \( a \): \( a = \frac{F_{\text{net}}}{m} \).
By completing these calculations, you can find the acceleration of the block, revealing how quickly it speeds up or slows down along the floor.

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Most popular questions from this chapter

A child weighing \(140 \mathrm{~N}\) sits at rest at the top of a playground slide that makes an angle of \(25^{\circ}\) with the horizontal. The child keeps from sliding by holding onto the sides of the slide. After letting go of the sides, the child has a constant acceleration of \(0.86 \mathrm{~m} / \mathrm{s}^{2}\) (down the slide, of course). (a) What is the coefficient of kinetic friction between the child and the slide? (b) What maximum and minimum values for the coefficient of static friction between the child and the slide are consistent with the information given here?

A student wants to determine the coefficients of static friction and kinetic friction between a box and a plank. She places the box on the plank and gradually raises one end of the plank. When the angle of inclination with the horizontal reaches \(30^{\circ}\), the box starts to slip, and it then slides \(2.5 \mathrm{~m}\) down the plank in \(4.0 \mathrm{~s}\) at constant acceleration. What are (a) the coefficient of static friction and (b) the coefficient of kinetic friction between the box and the plank?

A ski that is placed on snow will stick to the snow. However, when the ski is moved along the snow, the rubbing warms and partially melts the snow, reducing the coefficient of kinetic friction and promoting sliding. Waxing the ski makes it water repellent and reduces friction with the resulting layer of water. A magazine reports that a new type of plastic ski is especially water repellent and that, on a gentle \(200 \mathrm{~m}\) slope in the Alps, a skier reduced his top-to-bottom time from \(61 \mathrm{~s}\) with standard skis to \(42 \mathrm{~s}\) with the new skis. Determine the magnitude of his average acceleration with (a) the standard skis and (b) the new skis. Assuming a \(3.0^{\circ}\) slope, compute the coefficient of kinetic friction for (c) the standard skis and (d) the new skis.

A \(100 \mathrm{~N}\) force, directed at an angle \(\theta\) above a horizontal floor, is applied to a \(25.0 \mathrm{~kg}\) chair sitting on the floor. If \(\theta=0^{\circ}\), what are (a) the horizontal component \(F_{h}\) of the applied force and (b) the magnitude \(F_{N}\) of the normal force of the floor on the chair? If \(\theta=30.0^{\circ}\), what are (c) \(F_{h}\) and \((\) d \() F_{N}\) ? If \(\theta=60.0^{\circ}\), what are (e) \(F_{h}\) and (f) \(F_{N}\) ? Now assume that the coefficient of static friction between chair and floor is \(0.420\). Does the chair slide or remain at rest if \(\theta\) is \((\mathrm{g}) 0^{\circ},(\mathrm{h}) 30.0^{\circ}\), and (i) \(60.0^{\circ} ?\)

ao A \(2.5 \mathrm{~kg}\) block is initially at rest on a horizontal surface. A horizontal force \(\vec{F}\) of magnitude \(6.0 \mathrm{~N}\) and a vertical force \(\vec{P}\) are then applied to the block (Fig. \(6-17\) ). The coefficients of friction for the block and surface are \(\mu_{s}=0.40\) and \(\mu_{k}=0.25 .\) Determine the magnitude of the frictional force acting on the block if the magnitude of \(\vec{P}\) is (a) \(8.0 \mathrm{~N}\),(b) \(10 \mathrm{~N}\), and \((\mathrm{c}) 12 \mathrm{~N}\)
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