/*! 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 90 A girl is sledding down a slope ... [FREE SOLUTION] | 91Ó°ÊÓ

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A girl is sledding down a slope that is inclined at \(30.0^{\circ}\) with respect to the horizontal. The wind is aiding the motion by providing a steady force of \(105 \mathrm{N}\) that is parallel to the motion of the sled. The combined mass of the girl and the sled is \(65.0 \mathrm{kg}\), and the coefficient of kinetic friction between the snow and the runners of the sled is \(0.150 .\) How much time is required for the sled to travel down a \(175-\mathrm{m}\) slope, starting from rest?

Short Answer

Expert verified
It takes the sled approximately 8.17 seconds to travel down the slope.

Step by step solution

01

Identify the Forces Acting on the Sled

To solve the problem, we first identify all the forces acting on the sled. These include the gravitational force, the force of friction, and the additional force from the wind.- **Gravitational force**: This can be calculated using \( F_g = mg \sin(\theta) \) where \( m = 65.0 \mathrm{kg} \), \( g = 9.81 \mathrm{m/s^2} \), and \( \theta = 30.0^{\circ} \).- **Friction force**: This is \( F_{friction} = \mu_k \times N \), where \( \mu_k = 0.150 \) and \( N = mg \cos(\theta) \) is the normal force.- **Wind force**: Given as \( F_w = 105 \mathrm{N} \).
02

Calculate Gravitational Component Along Slope

Using the formula \( F_g = mg \sin(\theta) \), calculate the force due to gravity along the slope.\[ F_g = 65.0 \times 9.81 \times \sin(30.0^{\circ}) = 318.3 \mathrm{N} \]
03

Calculate Frictional Force

Calculate the frictional force using the formula \( F_{friction} = \mu_k \times N \) where the normal force \( N = mg \cos(\theta) \).\[ N = 65.0 \times 9.81 \times \cos(30.0^{\circ}) = 552.64 \mathrm{N} \]\[ F_{friction} = 0.150 \times 552.64 = 82.896 \mathrm{N} \]
04

Calculate Net Force

The net force on the sled is the sum of the forces along the direction of motion calculated by:\[ F_{net} = F_g + F_w - F_{friction} \]Substituting the values:\[ F_{net} = 318.3 + 105 - 82.896 = 340.404 \mathrm{N} \]
05

Calculate Acceleration

To find acceleration, use Newton’s second law \( F = ma \), where \( F \) is the net force and \( a \) is the acceleration.\[ 340.404 = 65.0 \times a \]\[ a = \frac{340.404}{65.0} = 5.237 \mathrm{m/s^2} \]
06

Calculate Time Using Kinematic Equation

Use the kinematic equation \( s = ut + \frac{1}{2}at^2 \) to find the time, where \( s \) is the distance (175 m), \( u \) is the initial velocity (0 m/s), and \( a \) is the acceleration.\[ 175 = 0 \times t + \frac{1}{2} \times 5.237 \times t^2 \]\[ t^2 = \frac{175 \times 2}{5.237} \]\[ t^2 = 66.800 \]\[ t = \sqrt{66.800} \approx 8.172 \mathrm{s} \]

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Key Concepts

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

Kinetic Friction
Kinetic friction is a force that opposes the motion of two surfaces sliding past each other. It acts parallel to the surfaces in contact and is always opposite to the direction of motion.
In the problem of sledding down the slope, the sled encounters kinetic friction between its runners and the snow.
This force can be calculated using the formula:
  • \( F_{\text{friction}} = \mu_k \times N \)
Here, \( \mu_k \) is the coefficient of kinetic friction, which tells us how "sticky" or "slippery" the surfaces are. Lower values mean less friction.
The normal force \( N \) is the component of the gravitational force perpendicular to the inclined plane. It can be found using:
  • \( N = mg \cos(\theta) \)
where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the inclined plane.
This frictional force works against the sled's motion, making it harder for the sled to accelerate down the slope.
Inclined Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. It is one of the basic machines identified in classical mechanics.
The main feature of an inclined plane is that it allows objects to slide or roll down more easily if the angle of inclination is increased.
In the sledding example, the plane is tilted at \(30^{\circ}\), which means that gravity not only pulls the sled downward but also helps propel it along the slope.
The force of gravity acting on an inclined plane can be broken down into two components:
  • \( F_{\text{parallel}} = mg \sin(\theta) \) (parallel to the slope, aiding motion)
  • \( F_{\text{perpendicular}} = mg \cos(\theta) \) (perpendicular to the slope, aiding frictional resistance)
By understanding these components, one can determine how the force of gravity helps or hinders movement on the inclined plane.
This understanding is crucial to solving any problem related to motion along an incline with forces such as friction and external aids like wind.
Net Force
Net force is the overall force acting on an object, resulting from the combination of all individual forces acting on it.
In the sled scenario, the net force dictates how fast the sled accelerates down the slope.
It is calculated using the equation:
  • \( F_{\text{net}} = F_{\text{gravity}} + F_{\text{wind}} - F_{\text{friction}} \)
Each term in this equation represents a distinct force:
  • \( F_{\text{gravity}} \) propels the sled downwards.
  • \( F_{\text{wind}} \) adds extra push in the sled's motion direction.
  • \( F_{\text{friction}} \) opposes motion, slowing it down.
Calculating these forces and combining them can help us use Newton's second law of motion \( F = ma \) to find the acceleration \( a \), which in turn can determine how quickly or slowly the sled descends the hill.
Net force not only describes motion but also the effect of multiple forces interacting, which underpin dynamics on any inclined plane like the sled example.

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