/*! 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 12 A \(1200-\mathrm{kg}\) car is be... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(1200-\mathrm{kg}\) car is being driven up a \(5.0^{\circ}\) hill. The frictional force is directed opposite to the motion of the car and has a magnitude of \(f=524 \mathrm{N}\) . A force \(\overrightarrow{\mathbf{F}}\) is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight \(\overrightarrow{\mathbf{W}}\) and the normal force \(\overrightarrow{\mathbf{F}}_{\mathrm{N}}\) directed perpendicular to the road surface. The length of the road up the hill is 290 \(\mathrm{m}\) . What should be the magnitude of \(\overrightarrow{\mathbf{F}},\) so that the net work done by all the forces acting on the car is \(+150 \mathrm{kJ}\) ?

Short Answer

Expert verified
The magnitude of the force \( \overrightarrow{\mathbf{F}} \) should be approximately 2130 N.

Step by step solution

01

Determine the gravitational force components

The weight of the car, \( \overrightarrow{\mathbf{W}} \), can be calculated using \( W = mg \), where \( m = 1200 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). Thus, \( W = 1200 \times 9.8 = 11760 \, \text{N} \). The component of this weight acting along the hill is \( W_{\parallel} = W \sin(\theta) \), where \( \theta = 5.0^{\circ} \). Thus, \( W_{\parallel} = 11760 \times \sin(5.0^{\circ}) \approx 1023.4 \, \text{N} \).
02

Identify the forces doing work

The forces that do work as the car moves up the hill are the applied force \( \overrightarrow{\mathbf{F}} \), the frictional force \( f = 524 \, \text{N} \), and the component of the car's weight along the slope \( W_{\parallel} \approx 1023.4 \, \text{N} \).
03

Calculate the net work required

The net work done by all forces is given as \( +150 \, \text{kJ} = 150,000 \, \text{J} \). This is the work needed to be achieved after considering the work done by friction and the gravitational component.
04

Set up the work-energy equation

The work done by the applied force \( \overrightarrow{\mathbf{F}} \) over the distance \( d = 290 \, \text{m} \) should counter the negative work done by friction and gravitational force, and achieve the desired net work: \[ F \cdot d - f \cdot d - W_{\parallel} \cdot d = 150,000 \] Substituting the known values gives: \[ F \cdot 290 - 524 \cdot 290 - 1023.4 \cdot 290 = 150,000 \]
05

Solve for the applied force \( F \)

Rearrange the equation from Step 4 to solve for \( F \): \[ F = \frac{150,000 + (524 \times 290) + (1023.4 \times 290)}{290} \] Calculate each term: \( 524 \times 290 = 151,960 \) and \( 1023.4 \times 290 = 296,786 \). Substitute back: \[ F = \frac{150,000 + 151,960 + 296,786}{290} \approx 2130 \] Therefore, the magnitude of the force \( \overrightarrow{\mathbf{F}} \) should be approximately \( 2130 \, \text{N} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Net Work
When various forces act on an object moving over a distance, the net work is the sum of the work done by each force. In mechanics, understanding net work is crucial because it tells us how the energy of an object changes. If the net work is positive, the object's kinetic energy increases. Conversely, if the net work is negative, the kinetic energy decreases.

In the exercise, the net work done on the car should be +150 kJ. This means the car's energy must increase by this amount, after balancing the effects of the forces acting against its motion, such as friction and the gravitational pull down the slope. Calculating net work involves summing contributions:
  • Applied force by the road.
  • Frictional force hindering the car.
  • Gravitational component pulling downward.
Ultimately, achieving the desired net work requires careful calculation to counteract these opposing forces and achieve the net positive energy change specified in the problem.
Frictional Force
Frictional force is a resistive force that opposes the relative motion of two surfaces in contact. It acts parallel to the surfaces and opposite to the direction of motion. In our exercise, the frictional force is given as 524 N, acting against the car's forward movement.

The role of friction in mechanics is central – it often resists movement, converting kinetic energy into thermal energy. This conversion causes a reduction in the net work done by other forces, which helps understand why the applied force needs to be greater than it seems.

To find the necessary force to move the car uphill effectively, this frictional force is subtracted from the work done. Thus, to overcome this resistance and achieve extra net work, the applied force must exert more energy than the frictional force dissipates. Remember, friction can be a complex variable as it can change with the surface condition and materials in contact.
Gravitational Force
Gravitational force, often referred to simply as weight, is the force with which Earth attracts a mass. Calculating gravitational force involves multiplying the mass of an object by the acceleration due to gravity (\( g = 9.8 \, \text{m/s}^2 \)).

For the car moving uphill, this force can be broken down into components parallel and perpendicular to the incline. The critical part here is the parallel component, which acts to pull the car back down the hill. This component is calculated using \( W_{\parallel} = W \sin(\theta) \), where \( \theta \) is the angle of the incline.
  • Calculating this helps determine how much force is needed to overcome gravity's pull.
Understanding gravitational forces in mechanics is key, especially in inclined plane problems, as it affects the net work required to move objects against this natural force. By accounting for gravitational work properly, one ensures that the applied forces suffice to provide the needed energy increase or sustained speed in the given direction.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A projectile of mass 0.750 kg is shot straight up with an initial speed of 18.0 m/s. (a) How high would it go if there were no air resistance? (b) If the projectile rises to a maximum height of only 11.8 m, determine the magnitude of the average force due to air resistance.

The surfer in the photo is catching a wave. Suppose she starts at the top of the wave with a speed of 1.4 m/s and moves down the wave until her speed increases to 9.5 m/s. The drop in her vertical height is 2.7 m. If her mass is 59 kg, how much work is done by the (non conservative) force of the wave?

The concepts in this problem are similar to those in Multiple-Concept Example 4, except that the force doing the work in this problem is the tension in the cable. A rescue helicopter lifts a 79-kg person straight up by means of a cable. The person has an upward acceleration of 0.70 \(m /s^{2}\) and is lifted from rest through a distance of 11 m. (a) What is the tension in the cable? How much work is done by (b) the tension in the cable and (c) the person’s weight? (d) Use the work–energy theorem and find the final speed of the person.

A 67.0-kg person jumps from rest off a 3.00-m-high tower straight down into the water. Neglect air resistance. She comes to rest 1.10 m under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is non conservative.

A gymnast is swinging on a high bar. The distance between his waist and the bar is 1.1 m, as the drawing shows. At the top of the swing his speed is momentarily zero. Ignoring friction and treating the gymnast as if all of his mass is located at his waist, find his speed at the bottom of the swing.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.