/*! 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 26 A \(3.5 \mathrm{~kg}\) block is ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 26

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. 6-49). 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 acceleration of the block.

Short Answer

Expert verified
(a) Frictional force: 6.17 N (b) Acceleration: 1.52 m/s^2

Step by step solution

01

- Determine Normal Force

First, calculate the normal force. Start by breaking the force \(\theta\) into components. \( F_y = F \sin(\theta) = 15 \sin(40^{\circ}) \approx 9.64 \text{N} \)The normal force \(F_N\) includes the gravitational force minus the vertical component of the applied force:\( F_N = mg - F_y = 3.5 \times 9.8 - 9.64 \approx 24.66 \text{N} \)
02

- Calculate Frictional Force

The frictional force \(f_k\) is given by the product of the normal force and the coefficient of kinetic friction:\( f_k = \mu_k F_N = 0.25 \times 24.66 \approx 6.17 \text{N} \)
03

- Determine Horizontal Component of Applied Force

Calculate the horizontal component of the applied force:\( F_x = F \cos(40^{\circ}) = 15 \cos(40^{\circ}) \approx 11.49 \text{N} \)
04

- Calculate Net Force

Determine the net force by subtracting the frictional force from the horizontal component of the applied force:\( F_{net} = F_x - f_k = 11.49 - 6.17 \approx 5.32 \text{N} \)
05

- Calculate Acceleration

Finally, use Newton鈥檚 second law to find the acceleration. \( F = ma \Rightarrow a = \frac{F_{net}}{m} = \frac{5.32}{3.5} \approx 1.52 \mathrm{m/s^2} \)

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Key Concepts

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

Normal Force
Normal force is the force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface. In our example, we need to consider both the gravitational force and the vertical component of the pushing force. The gravitational force acting on the block is given by \(mg\), where \(m = 3.5 \text{ kg}\) and \(g = 9.8 \text{ m/s}^2\). First, we resolve the pushing force into its vertical component: \( F_y = F \sin(40^\true) = 15 \sin(40^\true) \approx 9.64 \text{ N}\).

The normal force then becomes \(F_N = mg - F_y \approx 3.5 \times 9.8 - 9.64 \approx 24.66 \text{ N}\). This ensures the block remains in equilibrium vertically.
Kinetic Friction
Kinetic friction is the resisting force that acts opposite to the direction of motion when two objects are sliding past each other. The magnitude of the kinetic friction force can be calculated using the formula \( f_k = \mu_k F_N \), where \(\mu_k \) is the coefficient of kinetic friction, and \( F_N \) is the normal force. In our problem, \(\mu_k = 0.25\) and \( F_N \approx 24.66 \text{ N}\).

Therefore, the kinetic friction force is \( f_k = 0.25 \times 24.66 \approx 6.17 \text{ N}\). This force opposes the motion of the block on the floor.
Newton's Second Law
Newton's second law states that the sum of the forces acting on an object is equal to the mass of the object multiplied by its acceleration: \( \vec{F} = m \vec{a} \). This law helps us find the acceleration of the block by knowing the net force acting horizontally. To use this law, we first need to determine all the forces acting on the block, particularly in the horizontal direction. By Newton's second law, the net force \( F_{net} \) can be calculated and then used to find the acceleration using \( a = \frac{F_{net}}{m}\).

This principle guides us through determining the motion characteristics of the block once we have solved for all the forces involved.
Force Components
Understanding force components is crucial for solving problems involving angles. Forces can be split into horizontal and vertical components using trigonometric functions. For the horizontal component \( F_x\) of the applied force, we use cosine: \( F_x = F \cos(40^\true) \approx 15 \cos(40^\true) \approx 11.49 \text{ N}\).

These components allow us to analyze the effects of the applied force in specific directions. Ultimately, this separation simplifies calculations when determining other forces and the resulting motion. By breaking down forces, complex motion problems become more manageable.
Acceleration Calculation
Acceleration is calculated by finding the net force acting on an object and using Newton's second law. We've already calculated the components and forces acting against each other. The net force \( F_{net} \) is the difference between the horizontal component of the applied force and the frictional force: \( F_{net} = F_x - f_k = 11.49 - 6.17 \approx 5.32 \text{ N}\).

To find the acceleration, we apply Newton's second law: \( a = \frac{F_{net}}{m} = \frac{5.32 \text{ N}}{3.5 \text{ kg}} \approx 1.52 \text{ m/s}^2 \). This result indicates how quickly the block speeds up along the floor.

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Most popular questions from this chapter

A \(40 \mathrm{~kg}\) slab rests on a frictionless floor. A \(10 \mathrm{~kg}\) block rests on top of the slab (Fig. 6-58). The coefficient of static friction \(\mu^{\text {stat }}\) between the block and the slab is \(0.60\), whereas their kinetic friction coefficient \(\mu^{\text {kin }}\) is \(0.40\). The \(10 \mathrm{~kg}\) block is pulled by a horizontal force with a magnitude of \(100 \mathrm{~N}\). What are the resulting accelerations of (a) the block and (b) the slab?

We know that as an object passes through the air, the air exerts a resistive force on it. Suppose we have a spherical object of radius \(R\) and mass \(m\). What might the force plausibly depend on? \- It might depend on the properties of the object. The only ones that seem relevant are \(m\) and \(R\). \- It might depend on the object's coordinate and its derivatives: \(\vec{r}, \vec{v}, \vec{a}, \ldots\) \- It might depend on the properties of the air, such as the density, \(\rho\). (a) Explain why it is plausible that the force the air exerts on a sphere depends on \(R\) but implausible that it depends on \(m\). (b) Explain why it is plausible that the force the air exerts depends on the object's speed through it, \(|\vec{v}|\), but not on its position, \(\vec{r}\), or acceleration, \(\vec{a}\). (c) Dimensional analysis is the use of units (e.g., meters, seconds, or newtons) associated with quantities to reason about the relationship between the quantities. Using dimensional analysis, construct a plausible form for the force that air exerts on a spherical body moving through it.

A bedroom bureau with a mass of \(45 \mathrm{~kg}\). including drawers and clothing, rests on the floor. (a) If the coefficient of static friction between the bureau and the floor is \(0.45\), what is the magnitude of the minimum horizontal force that a person must apply to start the bureau moving? (b) If the drawers and clothing, with \(17 \mathrm{~kg}\) mass, are removed before the bureau is pushed, what is the new minimum magnitude?

A \(68 \mathrm{~kg}\) crate is dragged across a floor by pulling on a rope attached to the crate and inclined \(15^{\circ}\) above the horizontal. (a) If the coefficient of static friction is \(0.50\), what minimum force magnitude is required from the rope to start the crate moving? (b) If \(\mu^{\text {kin }}\) \(=0.35\), what is the magnitude of the initial acceleration of the crate?

The coefficient of static friction between Teflon and scrambled eggs is about \(0.04 .\) What is the smallest angle from the horizontal that will cause the eggs to slide across the bottom of a Teflon-coated skillet?
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