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A child whose weight is \(267 \mathrm{~N}\) slides down a \(6.1 \mathrm{~m}\) playground slide that makes an angle of \(20^{\circ}\) with the horizontal. The coefficient of kinetic friction between slide and child is 0.10 . (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of \(0.457 \mathrm{~m} / \mathrm{s},\) what is her speed at the bottom?

Short Answer

Expert verified
(a) 15.54 J (b) 4.01 m/s

Step by step solution

01

Calculate the Gravitational Force Component Along the Slide

Decompose the gravitational force into components. The force along the slide is given by \( F_{\parallel} = mg \sin \theta \), where \( m \) is the mass (\( m = \frac{267\, \text{N}}{9.8\, \text{m/s}^2} \)), \( g \) is the acceleration due to gravity \( (9.8 \text{ m/s}^2) \), and \( \theta \) is the angle of the slide (\( 20^{\circ} \)). Substitute the values and solve for \( F_{\parallel} \).
02

Calculate the Normal Force

Calculate the normal force, which is perpendicular to the slide surface, using the formula \( F_{\perp} = mg \cos \theta \). Using the given weight \( mg = 267 \text{ N} \) and angle \( \theta = 20^{\circ} \), substitute to find \( F_{\perp} \).
03

Calculate the Force of Friction

The force of friction is calculated using \( f_k = \mu_k F_{\perp} \). Here, \( \mu_k = 0.10 \) is the coefficient of kinetic friction. Substitute the normal force from Step 2 to get \( f_k \).
04

Calculate the Work Done by Friction (Thermal Energy)

The work done by friction, which equals the energy transferred to thermal energy, is given by \( W = f_k \cdot d \), where \( d = 6.1 \text{ m} \) is the distance of the slide. Use the frictional force from Step 3 to calculate \( W \).
05

Apply the Conservation of Energy for Final Speed

Use the conservation of energy: \( E_i + W = E_f \). Initial energy \( E_i = \frac{1}{2}mv_i^2 + mgh \), and final energy \( E_f = \frac{1}{2}mv_f^2 \). The height \( h \) is found using \( h = d \sin \theta \). Substitute \( v_i = 0.457 \text{ m/s} \), solve for \( v_f \).

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Key Concepts

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

Gravitational Force
The gravitational force is a fundamental force that pulls objects towards the Earth. When a child slides down a playground slide, gravity plays a crucial role in determining the motion. The gravitational force acting on the child can be decomposed into two components: one parallel to the slide and the other perpendicular to it.
The force along the slide, which actually helps the child slide down, is known as the gravitational force component along the slide. It can be calculated using the equation:
  • \( F_{\parallel} = mg \sin \theta \)
Here, \( m \) stands for mass, \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \)), and \( \theta \) is the angle of the slide. For example, if a child weighs 267 N and the slide angle is 20°, we can calculate the force component responsible for the motion.
Understanding this component is key to solving problems in mechanics, like finding the net forces and motion on an inclined plane.
Energy Conservation
The principle of energy conservation is a crucial concept in physics, especially when analyzing motion involving gravitational forces and friction. This principle states that the total mechanical energy in a system remains constant, provided no external forces do work. However, in real-world scenarios, energy is transferred between types, like from kinetic to thermal due to friction.
When the child slides down the slide, her initial energy is a combination of her gravitational potential energy and kinetic energy. As she descends, the potential energy decreases while her kinetic energy increases, allowing her to speed up. Nevertheless, some of this energy is converted into thermal energy, due to friction, which we need to account for in energy calculations.
Using energy conservation, we write:
  • \( E_i + W = E_f \)
  • Where \( E_i \) is initial energy, \( W \) is work done by friction, and \( E_f \) is final energy.
This equation is pivotal in solving problems related to motion on slides or inclined planes, helping us determine final speeds or energy transformations.
Work Done by Friction
Friction is an opposing force that resists the motion of objects. In the context of the slide, kinetic friction comes into play, opposing the child's motion and converting kinetic energy into thermal energy along the slide. This conversion is why the slide feels warm after repeated use.
The work done by friction, or the energy transferred to thermal energy, can be calculated using the formula:
  • \( W = f_k \cdot d \)
Where \( f_k \) is the kinetic friction force and \( d \) is the distance the child slides. The kinetic friction force is given by \( f_k = \mu_k F_\perp \), where \( \mu_k \) is the coefficient of kinetic friction and \( F_\perp \) is the normal force. These calculations not only enable us to understand energy losses in mechanical systems but also demonstrate real-life applications of physics in everyday activities.

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Most popular questions from this chapter

The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$ U=\frac{A}{r^{12}}-\frac{B}{r^{6}} $$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the \(e q u i-\) librium separation \(-\) that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

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110 A \(5.0 \mathrm{~kg}\) block is projected at \(5.0 \mathrm{~m} / \mathrm{s}\) up a plane that is inclined at \(30^{\circ}\) with the horizontal. How far up along the plane does the block go (a) if the plane is frictionless and (b) if the coefficient of kinetic friction between the block and the plane is \(0.40 ?\) (c) In the latter case, what is the increase in thermal energy of block and plane during the block's ascent? (d) If the block then slides back down against the frictional force, what is the block's speed when it reaches the original projection point?

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