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Chapter 6: Problem 14

You apply a constant force \(\overrightarrow{F} =(-68.0 \, \mathrm{N})\hat{\imath} +(36.0 \, \mathrm{N})\hat{\jmath}\) to a 380-kg car as the car travels 48.0 m in a direction that is 240.0\(^\circ\) counterclockwise from the +\(x\)-axis. How much work does the force you apply do on the car?

Short Answer

Expert verified
The work done on the car is 135.552 J.

Step by step solution

01

Understand the concept of work

The work done (W) by a force is given by the dot product of the force vector \( \overrightarrow{F} \) and the displacement vector \( \overrightarrow{s} \). The formula for work is: \[ W = \overrightarrow{F} \cdot \overrightarrow{s} = F_x s_x + F_y s_y \] where \( F_x \) and \( F_y \) are the components of the force, and \( s_x \) and \( s_y \) are the components of the displacement.
02

Calculate displacement components

The car travels in a direction of 240.0掳 counterclockwise from the +x-axis. Break the displacement vector \( \overrightarrow{s} \) into its x and y components. The displacement is 48.0 m long. Use trigonometric functions: \[ s_x = 48.0 \cos(240.0^{\circ}) \]\[ s_y = 48.0 \sin(240.0^{\circ}) \] Calculating gives: \( s_x = 48.0 (-0.5) = -24.0 \, \text{m} \) and \( s_y = 48.0 (-0.866) = -41.568 \, \text{m} \).
03

Use the work formula

Substitute the components of force and displacement into the work formula: \[ W = (-68.0 \, \text{N})(-24.0 \, \text{m}) + (36.0 \, \text{N})(-41.568 \, \text{m}) \] Calculating gives: \( W = 1632.0 \, \text{N}\cdot\text{m} - 1496.448 \, \text{N}\cdot\text{m} = 135.552 \, \text{J} \).
04

State the amount of work done

Thus, the work done by the force \( \overrightarrow{F} \) on the car as it travels 48.0 m is 135.552 J.

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Key Concepts

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

Force Vector
A force vector is a way to express the magnitude and direction of a force applied to an object. In the exercise, the force vector \( \overrightarrow{F} \) is given as \((-68.0 \, \mathrm{N})\hat{\imath} + (36.0 \, \mathrm{N})\hat{\jmath}\). This notation is called unit vector notation, where:
  • \(\hat{\imath}\) represents the component in the x-direction.
  • \(\hat{\jmath}\) represents the component in the y-direction.
Each of these components shows how much force is applied along the respective axis. The negative sign for the x-component indicates the direction of the force is opposite to the positive x-direction. Understanding force vectors is crucial in determining how forces affect the movement of objects in physics.
Displacement Vector
The displacement vector represents how far and in what direction an object moves from its initial position. In the context of this exercise, the car's movement is described by the displacement vector \( \overrightarrow{s} \). The car travels a distance of 48.0 m in a direction specified by an angle of 240.0掳 counterclockwise from the positive x-axis.
The displacement vector can be broken into two components, using trigonometric functions:
  • The x-component \( s_x = 48.0 \cos(240.0^{\circ}) \)
  • The y-component \( s_y = 48.0 \sin(240.0^{\circ}) \)
These calculations use the cosine and sine of the given angle to find the respective components, showing how the total displacement divides between horizontal and vertical directions.
Dot Product
The dot product is a mathematical operation that takes two equal-length sequences of numbers (usually vectors) and returns a single number. In physics, this is used to calculate work done by a force, as work is the dot product of the force vector \( \overrightarrow{F} \) and the displacement vector \( \overrightarrow{s} \). The formula for work is:
\[ W = \overrightarrow{F} \cdot \overrightarrow{s} = F_x s_x + F_y s_y \]
  • \( F_x \) and \( F_y \) are the components of the force.
  • \( s_x \) and \( s_y \) are the components of the displacement.
The dot product can tell us how much of the force is used in doing work in the direction of the displacement. It's important in calculating realistic scenarios where forces do not act in the same direction as movement.
Trigonometry in Physics
Trigonometry is a branch of mathematics that studies the relations between side lengths and angles of triangles. In physics, it is often used to resolve vectors, like forces and displacements, into their components.
In this exercise, to find the components of the displacement vector \( \overrightarrow{s} \), we utilize basic trigonometric functions:
  • \(\cos\) to find the horizontal component \( s_x \)
  • \(\sin\) to find the vertical component \( s_y \)
The angle provided (240.0 degrees) is used in conjunction with these functions to determine how the 48.0 m displacement splits into horizontal and vertical parts. Understanding how to apply trigonometry in this context is crucial for solving many problems involving vectors in physics.

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Most popular questions from this chapter

A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest on a horizontal floor. It is then pushed in a straight line for 1.20 m by a trained dog that exerts a horizontal force with magnitude 36.0 N. Use the work\(-\)energy theorem to find the final speed of the 12-pack if (a) there is no friction between the 12-pack and the floor, and (b) the coefficient of kinetic friction between the 12-pack and the floor is 0.30.

Use the work\(-\)energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases. (a) A branch falls from the top of a 95.0-m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 m into the air. How fast was the boulder moving just as it left the volcano?

You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground, it is traveling at 25.0 m/s upward. Use the work\(-\)energy theorem to find (a) the rock's speed just as it left the ground and (b) its maximum height.

We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass \(M\), equilibrium length \(L_0\), and force constant \(k\). The work done to stretch or compress the spring by a distance \(L\) is \\(\frac{1}{2}\\) \(kX^2\), where \(X = L - L_0\). Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\). Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. (a) Calculate the kinetic energy of the spring in terms of \(M\) and \(v\). (\(Hint\): Divide the spring into pieces of length \(dl\); find the speed of each piece in terms of \(l\), \(v\), and \(L\); find the mass of each piece in terms of \(dl\), \(M\), and \(L\); and integrate from \(0\) to \(L\). The result is \(not\) \\(\frac{1}{2}\\) \(Mv^2\), since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 kg and force constant 3200 N/m is compressed 2.50 cm from its unstretched length. When the trigger is pulled, the spring pushes horizontally on a 0.053-kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncompressed length (b) ignoring the mass of the spring and (c) including, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

(a) In the Bohr model of the atom, the ground state electron in hydrogen has an orbital speed of 2190 km/s. What is its kinetic energy? (Consult Appendix F.) (b) If you drop a 1.0-kg weight (about 2 lb) from a height of 1.0 m, how many joules of kinetic energy will it have when it reaches the ground? (c) Is it reasonable that a 30-kg child could run fast enough to have 100 J of kinetic energy?
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