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Chapter 6: Problem 72

A child goes down a playground slide that is inclined at an angle of \(26.5^{\circ}\) below the horizontal. Find the acceleration of the child given that the coefficient of kinetic friction between the child and the slide is 0.315.

Short Answer

Expert verified
The child's acceleration is approximately 1.604 m/s虏.

Step by step solution

01

Understand the Forces Involved

Firstly, identify the forces acting on the child sliding down the playground slide. The forces involved are the gravitational force pulling the child downward, the normal force exerted perpendicular to the surface of the slide, and the kinetic frictional force opposing the motion.
02

Set Up the Free-Body Diagram

Create a free-body diagram of the forces. The gravitational force can be split into two components: one parallel to the slide (which causes the motion) and one perpendicular to the slide.
03

Calculate the Gravitational Force Components

The gravitational force acting parallel to the slide is given by \(mg\sin(\theta)\) where \(m\) is the mass of the child, \(g = 9.8 \, m/s^2\) is the gravitational acceleration, and \(\theta = 26.5^{\circ}\) is the angle of inclination. The force normal to the slide is \(mg\cos(\theta)\).
04

Determine the Frictional Force

The kinetic frictional force is calculated by multiplying the normal force by the coefficient of kinetic friction. So, \(f_{friction} = \mu_k \cdot mg\cos(\theta)\) where \(\mu_k = 0.315\) is the coefficient of friction.
05

Apply Newton's Second Law

Apply Newton's Second Law along the direction of the slide: \(mg\sin(\theta) - f_{friction} = ma\), where \(a\) is the acceleration.
06

Solve for the Acceleration

Rearrange the equation to solve for \(a\): \[ a = g\sin(\theta) - \mu_k g\cos(\theta) \]Substitute the known values: \[ a = 9.8 \times \sin(26.5^{\circ}) - 0.315 \times 9.8 \times \cos(26.5^{\circ}) \].
07

Calculate the Numerical Value of Acceleration

Perform the calculations:\( \sin(26.5^{\circ}) \approx 0.446 \) and \( \cos(26.5^{\circ}) \approx 0.894 \).Therefore, \[ a = 9.8 \times 0.446 - 0.315 \times 9.8 \times 0.894 \] \[ a \approx 4.371 - 2.767 \approx 1.604 \, m/s^2 \].

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Key Concepts

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

Gravitational Force Components
When a child goes down a slide, the gravitational force that pulls them downward can be broken into two components. This is because the slide is inclined rather than vertical. Understanding these components helps in analyzing the forces at play.
  • Parallel Component: This is the part of the gravitational force that acts along the slide, making the child move downwards. It's calculated using the formula: \[mg\sin(\theta)\] where \(m\) is the mass of the child, \(g = 9.8\, m/s^2\) is the acceleration due to gravity, and \(\theta\) is the angle of the slide with the horizontal.
  • Perpendicular Component: This component acts perpendicular to the slide. It's what the surface of the slide "feels" from the child's weight, influencing the normal force. The formula to find it is: \[mg\cos(\theta)\] which helps in calculating kinetic friction as it affects how tightly the child is pressing against the slide.
By breaking the gravitational force into these components, we can get a clear picture of how each force affects the child's motion.
Newton's Second Law
Newton's Second Law is a fundamental principle that tells us how forces affect the motion of an object. In the context of the child on the slide, it helps us calculate the acceleration.
The law states: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The formula is written as: \[F = ma\] where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration.
  • Calculating Net Force: As the child slides down, the net force is the difference between the gravitational force component along the slide and the frictional force opposed to motion.
  • Application: By rearranging the formula to solve for \(a\), you get: \[a = \frac{F}{m} \] In our slide problem, the equation becomes: \[a = g\sin(\theta) - \mu_k g\cos(\theta)\] where \(\mu_k\) is the coefficient of kinetic friction.
Using Newton's Second Law, we convert all the forces into a single acceleration value. This explains how fast the child will move down the slide under the influence of given forces.
Free-Body Diagram
A free-body diagram is a simple and effective visual tool used in physics. It helps isolate and identify all the forces acting on an object. In this case, it allows us to see clearly what the child on the slide experiences.
  • Sketching the Forces: In the free-body diagram for our problem, you'd illustrate: - the weight of the child pulling them down due to gravity (with components along and perpendicular to the slide), - the normal force acting perpendicularly outwards from the slide, and - the frictional force opposing the downward slide direction.
  • Purpose: The diagram breaks down complex force interactions into more manageable parts. It avoids confusion and helps with the calculation of forces, especially when components like friction and angles are involved.
Using a free-body diagram is crucial. It ensures all forces are accounted for accurately, which leads to precise calculations of motion and force interactions.

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Most popular questions from this chapter

The equilibrium length of a certain spring with a force constant of \(k=250 \mathrm{N} / \mathrm{m}\) is \(0.18 \mathrm{m}\). (a) What is the magnitude of the force that is required to hold this spring at twice its equilibrium length? (b) Is the magnitude of the force required to keep the spring compressed to half its equilibrium length greater than, less than, or equal to the force found in part (a)? Explain.

A baseball player slides into third base with an initial speed of \(4.0 \mathrm{m} / \mathrm{s} .\) If the coefficient of kinetic friction between the player and the ground is \(0.46,\) how far does the player slide before coming to rest?

Two drivers traveling side-by-side at the same speed suddenly see a deer in the road ahead of them and begin braking. Driver 1 stops by locking up his brakes and screeching to a halt; driver 2 stops by applying her brakes just to the verge of locking, so that the wheels continue to turn until her car comes to a complete stop. (a) All other factors being equal, is the stopping distance of driver 1 greater than, less than, or equal to the stopping distance of driver \(2 ?\) (b) Choose the best explanation from among the following: I. Locking up the brakes gives the greatest possible braking force. II. The same tires on the same road result in the same force of friction. III. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction.

Find the coefficient of kinetic friction between a \(3.85-\mathrm{kg}\) block and the horizontal surface on which it rests if an \(850-\mathrm{N} / \mathrm{m}\) spring must be stretched by \(6.20 \mathrm{cm}\) to pull it with constant speed. Assume that the spring pulls in the horizontal direction.

As plants grow, they tend to align their stems and roots along the direction of the gravitational field. This tendency, which is related to differential concentrations of plant hormones known as auxins, is referred to as gravitropism. As an illustration of gravitropism, experiments show that seedlings placed in pots on the rim of a rotating turntable do not grow in the vertical direction. Do you expect their stems to tilt inward-toward the axis of rotation -or outward-away from the axis of rotation?
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