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Chapter 6: Problem 98

A child has mass 6.0 kg and slides down a 35" incline with constant speed under the action of a \(34-\mathrm{N}\) force acting up and parallel to the incline. What is the coefficient of kinetic friction between the child and the surface of the incline?

Short Answer

Expert verified
The coefficient of kinetic friction between the child and the surface of the incline is 0.0083.

Step by step solution

01

Identify the forces and draw a Free Body Diagram (FBD)

Draw a free body diagram of the child on the incline. The forces acting on the child are: 1. Gravitational force (weight) acting downwards: \(F_g = mg\) 2. Normal force acting perpendicular to the incline: \(F_N\) 3. Friction force acting up the incline: \(F_f\) 4. External force acting up and parallel to the incline: 34 N
02

Resolve forces into components

Resolve the gravitational force into components along the incline (in the horizontal direction) and perpendicular to the incline (in the vertical direction). The horizontal component of the gravitational force: \(F_{gx} = mg\sin(\theta)\), where \(\theta\) is the angle of the incline. The vertical component of the gravitational force: \(F_{gy} = mg\cos(\theta)\).
03

Apply Newton's second law

Since the child is moving at a constant velocity, the net force along the incline is zero. Therefore, we can write the equation of motion along the incline: \(F_f = F_{gx} - 34\) Since the child is not accelerating in the vertical direction, we can write the equation for the vertical direction: \(F_N = F_{gy}\)
04

Calculate friction force

We can first find the horizontal and vertical components of the gravitational force: \(F_{gx} = mg\sin(\theta) = (6.0\,\text{kg})\times(9.8\,\text{m/s}^2)\times\sin(35^\circ) = 33.6\,\text{N}\) \(F_{gy} = mg\cos(\theta) = (6.0\,\text{kg})\times(9.8\,\text{m/s}^2)\times\cos(35^\circ) = 48.0\,\text{N}\) Now, use the equation of motion along the incline to find the friction force: \(F_f = F_{gx} - 34 = 33.6\,\text{N} - 34\,\text{N} = -0.4\,\text{N}\) Since the friction force is acting opposite to the direction of the external force, we must consider the magnitude of the friction force, which is \(0.4\,\text{N}\).
05

Calculate the coefficient of kinetic friction

The equation for the friction force is: \(F_f = \mu_k F_N\) We already know the normal force, so we can find the coefficient of kinetic friction: \(\mu_k = \frac{F_f}{F_N} = \frac{0.4\,\text{N}}{48.0\,\text{N}} = 0.0083\) #Solution# The coefficient of kinetic friction between the child and the surface of the incline is 0.0083.

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Key Concepts

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

Free Body Diagram
Understanding the interactions of forces in a physical scenario is fundamental to solving problems in mechanics. One powerful method for visualizing these forces is by drawing a Free Body Diagram (FBD). An FBD represents all of the forces acting on a single object, with vectors indicating the direction of each force. It's crucial to keep the diagram simple and only include forces that directly impact the object's motion. For example, in the given exercise, the child sliding down the incline has several forces acting on them: the force of gravity, the normal force from the incline's surface, the friction force opposing the motion, and the external force applied. By drawing a Free Body Diagram, we accurately visualize and identify these forces, setting the stage for a successful application of Newton's laws of motion.
Newton's Second Law
At the heart of classical mechanics lies Newton's second law, stated as \( F = ma \), where \( F \) is the net force acting on a body, \( a \) is the acceleration, and \( m \) is the mass of the body. In cases where an object is moving at a consistent speed (constant velocity), the acceleration is zero, and hence the net force is also zero. This understanding is particularly crucial for our problem, as it implies that the sum of the forces parallel to the incline must cancel out, resulting in no net acceleration of the child down the incline. By applying Newton's second law, we can establish relationships between the known forces and use them to solve for unknown quantities, such as the coefficient of kinetic friction.
Friction Force
Friction is the force that opposes the relative motion of two surfaces in contact. It's often depicted in FBDs and is critical in many physical situations. The friction force depends on the nature of the surfaces in contact and the normal force pressing them together. There are two types of frictional forces relevant to physics problems: static friction, which acts when objects are stationary, and kinetic friction, which acts when objects are moving. In the exercise, the child is sliding at a constant speed down the incline, which means the kinetic friction force is at play. Understanding how to calculate this force is essential as it allows us to determine the coefficient of kinetic friction, ultimately helping us comprehend the interaction between the child and the surface of the incline.
Gravitational Force Components
When dealing with inclined planes, it's important to consider the gravitational force components acting on a body. Gravity always exerts a force directly toward the center of the Earth, but when an object is on an incline, this force can be resolved into two components: one perpendicular to the surface (responsible for the normal force) and one parallel to the surface (affecting the object's motion along the incline). The exercise involves calculating these components using trigonometric functions, specifically sine and cosine, based on the angle of the incline. By doing so, we can find the effective force trying to pull the child down the slide and the normal force, which is crucial in finding the coefficient of kinetic friction.

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Most popular questions from this chapter

A small space probe is released from a spaceship. The space probe has mass \(20.0 \mathrm{kg}\) and contains \(90.0 \mathrm{kg}\) of fuel. It starts from rest in deep space, from the origin of a coordinate system based on the spaceship, and burns fuel at the rate of \(3.00 \mathrm{kg} / \mathrm{s}\). The engine provides a constant thrust of \(120.0 \mathrm{N}\). (a) Write an expression for the mass of the space probe as a function of time, between 0 and 30 seconds, assuming that the engine ignites fuel beginning at \(t=0 .\) (b) What is the velocity after \(15.0 \mathrm{s}\) ? (c) What is the position of the space probe after \(15.0 \mathrm{s}\), with initial position at the origin? (d) Write an expression for the position as a function of time, for \(t>30.0 \mathrm{s}\)

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