/*! 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 91 A motorcycle and \(60.0 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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A motorcycle and \(60.0 \mathrm{~kg}\) rider accelerate at \(3.0 \mathrm{~m} / \mathrm{s}^{2}\) up a ramp inclined \(10^{\circ}\) above the horizontal. What are the magnitudes of (a) the net force on the rider and (b) the force on the rider from the motorcycle?

Short Answer

Expert verified
(a) Net force on the rider is 180 N. (b) Force by motorcycle on the rider is 282.41 N.

Step by step solution

01

Identify Given Information

First, let's identify the information given in the problem. We have the mass of the rider, \(m = 60.0 \text{ kg}\), and the acceleration, \(a = 3.0 \text{ m/s}^2\). The ramp is inclined at \(10^\circ\).
02

Calculate the Net Force on the Rider

The net force can be calculated using Newton's second law: \(F_{\text{net}} = ma\). Substituting the given values, we have \(F_{\text{net}} = 60.0 \text{ kg} \times 3.0 \text{ m/s}^2 = 180.0 \text{ N}\).
03

Determine the Components of Forces on the Inclined Plane

On an inclined plane, the forces acting on the rider include gravitational force, friction (which is zero or negligible in ideal cases), and the force from the motorcycle. The gravitational force has a component parallel to the incline: \(F_{g, \parallel} = mg \sin(\theta)\), where \(g = 9.8 \text{ m/s}^2\) and \(\theta = 10^\circ\).
04

Calculate the Gravitational Force Parallel to the Incline

Calculate \(F_{g, \parallel} = 60.0 \text{ kg} \times 9.8 \text{ m/s}^2 \times \sin(10^\circ)\). First, calculate \(\sin(10^\circ) \approx 0.1736\). Then, \(F_{g, \parallel} \approx 60.0 \times 9.8 \times 0.1736 = 102.41 \text{ N}\).
05

Calculate the Force by Motorcycle on Rider

The force exerted by the motorcycle on the rider, \(F_m\), is what supplies both the net force and compensates for the component of gravitational force. Thus, \(F_m = F_{\text{net}} + F_{g, \parallel}\). Substituting the calculated values, \(F_m = 180.0 \text{ N} + 102.41 \text{ N} = 282.41 \text{ N}\).

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Key Concepts

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

Inclined Plane
An inclined plane is a flat surface that is tilted at an angle from the horizontal. It's one of the classic simple machines and has been utilized for centuries to make moving or lifting objects easier. When an object is placed on an incline, certain forces act upon it differently than they would on a flat surface. This is why the study of physics on inclined planes can often be quite engaging and enlightening.

In the context of this exercise, the motorcycle and rider are moving up a ramp inclined at 10 degrees. This incline affects the way forces are distributed and calculated. An incline reduces the effect of gravity's pull directly in opposition to motion, breaking gravity into components that run parallel and perpendicular to the surface.

Understanding how inclined planes work allows students to explore how different forces impact movement. It is essential to know how these forces can be calculated to determine net forces efficiently. Inclined planes transform how forces like gravity interact with objects, making it easier to lift or push objects. This explains why ramps are so commonly used in everyday life.
Gravitational Force
Gravitational force, often symbolized as \( F_g \), is the attractive force that Earth exerts on objects, pulling them towards its center. It is a crucial force that influences motion, especially on inclined planes, where it has a unique function. For an object on an incline, the gravitational force can be divided into two components: one that is perpendicular to the plane and another that is parallel.

The parallel component of gravitational force can be particularly important for understanding motion on ramps, and is calculated using the formula:
  • \( F_{g, \, \parallel} = mg \sin(\theta) \)
where:
  • \( m \) is the mass of the object
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \)
  • \( \theta \) is the angle of the incline
In our problem, the gravitational force plays a crucial role in determining the forces the motorcycle must overcome to accelerate upward. The inclination modifies how the gravitational force is applied, allowing us to use trigonometry to solve for specific force components. Understanding these components helps in accurately calculating the net force and understanding the mechanics of motion on an incline.
Net Force Calculation
Calculating the net force on an object is fundamental in understanding its motion. According to Newton's Second Law of Motion, the net force \( F_{\text{net}} \) acting on an object is equal to the product of its mass \( m \) and its acceleration \( a \). This relationship is expressed by the formula:
  • \( F_{\text{net}} = ma \)
In this exercise, the motorcycle's acceleration is up the ramp and is calculated as having a net force of \( 180.0 \, \text{N} \) acting on the rider. This net force is responsible for accelerating the rider and motorcycle against the gravitational pull and any other resistive forces.

The calculation of the net force must account for the incline and other forces, such as the gravitational force component. The calculated net force provides insight into the overall dynamics of motion. It also underlines the idea that inclines can modify how forces need to be applied to achieve a certain motion.

Where multiple forces are acting, such as gravitational components, the vector sum of these forces gives the net force. Solving this helps us understand how much additional force, such as that provided by the motorcycle, is required to maintain or change motion effectively.

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Most popular questions from this chapter

Tarzan, who weighs \(820 \mathrm{~N}\), swings from a cliff at the end of a \(20.0 \mathrm{~m}\) vine that hangs from a high tree limb and initially makes an angle of \(22.0^{\circ}\) with the vertical. Assume that an \(x\) axis extends horizontally away from the cliff edge and a \(y\) axis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is \(760 \mathrm{~N}\). Just then, what are (a) the force on him from the vine in unit- vector notation and the net force on him (b) in unit-vector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the \(x\) axis? What are the (e) magnitude and (f) angle of Tarzan's acceleration just then?

A \(2.00 \mathrm{~kg}\) object is subjected to three forces that give it an acceleration \(\vec{a}=-\left(8.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}} . \quad\) If two of the three forces are \(\vec{F}_{1}=(30.0 \mathrm{~N}) \hat{\mathrm{i}}+(16.0 \mathrm{~N}) \hat{\mathrm{j}}\) and \(\vec{F}_{2}=\) \(-(12.0 \mathrm{~N}) \hat{\mathrm{i}}+(8.00 \mathrm{~N}) \hat{\mathrm{j}},\) find the third force.

The Zacchini family was renowned for their human-cannonball act in which a family member was shot from a cannon using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon and at a range of \(69 \mathrm{~m} .\) He was propelled inside the barrel for \(5.2 \mathrm{~m}\) and launched at an angle of \(53^{\circ} .\) If his mass was \(85 \mathrm{~kg}\) and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at \(53^{\circ} .\) Neglect air drag.)

A lamp hangs vertically from a cord in a descending elevator that decelerates at \(2.4 \mathrm{~m} / \mathrm{s}^{2}\). (a) If the tension in the cord is \(89 \mathrm{~N},\) what is the lamp's mass? (b) What is the cord's tension when the elevator ascends with an upward acceleration of \(2.4 \mathrm{~m} / \mathrm{s}^{2} ?\)

A \(2.0 \mathrm{~kg}\) particle moves along an \(x\) axis, being propelled by a variable force directed along that axis. Its position is given by \(x=3.0 \mathrm{~m}+(4.0 \mathrm{~m} / \mathrm{s}) t+c t^{2}-\left(2.0 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3},\) with \(x\) in meters and \(t\) in seconds. The factor \(c\) is a constant. At \(t=3.0 \mathrm{~s},\) the force on the particle has a magnitude of \(36 \mathrm{~N}\) and is in the negative direction of the axis. What is \(c ?\)

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