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Chapter 7: Problem 71

A constant force of magnitude \(10 \mathrm{~N}\) makes an angle of \(150^{\circ}\) (measured counterclockwise) with the positive \(x\) direction as it acts on a \(2.0 \mathrm{~kg}\) object moving in an \(x y\) plane. How much work is done on the object by the force as the object moves from the origin to the point having position vector \((2.0 \mathrm{~m}) \hat{\mathrm{i}}-(4.0 \mathrm{~m}) \hat{\mathrm{j}}\) ?

Short Answer

Expert verified
The work done is \(-10\sqrt{3} - 20\) Joules.

Step by step solution

01

Identify the Components of the Force

The problem gives us a force with magnitude 10 N at an angle of 150掳 from the positive x direction. To find its components, use the formulas: \( F_x = F \cos(\theta) \) and \( F_y = F \sin(\theta) \). Thus, \( F_x = 10 \cos(150掳) \) and \( F_y = 10 \sin(150掳) \).
02

Calculate the Components of the Force

Using trigonometric values for 150掳, \( \cos(150掳) = -\frac{\sqrt{3}}{2} \) and \( \sin(150掳) = \frac{1}{2} \). Therefore, \( F_x = 10 \cdot -\frac{\sqrt{3}}{2} = -5\sqrt{3} \), and \( F_y = 10 \cdot \frac{1}{2} = 5 \).
03

Define the Displacement Vector

The object moves from the origin to position \( (2.0 \hat{i} - 4.0 \hat{j}) \). Hence, the displacement vector \( \vec{d} \) is given as \( \vec{d} = 2 \hat{i} - 4 \hat{j} \).
04

Compute Work Done Using Dot Product

Work done is calculated using the dot product of force and displacement vectors: \( W = \vec{F} \cdot \vec{d} = (F_x \hat{i} + F_y \hat{j}) \cdot (d_x \hat{i} + d_y \hat{j}) \). Substituting the values, \[ W = (-5\sqrt{3}) \cdot 2 + 5 \cdot (-4) \].
05

Evaluate the Dot Product

Calculate: \[ W = (-5\sqrt{3}) \cdot 2 + 5 \cdot (-4) = -10\sqrt{3} - 20 \].
06

Provide Final Work Done

The total work done by the force on the object, evaluated from the dot product, is \( -10\sqrt{3} - 20 \) Joules.

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Key Concepts

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

Force Components
When dealing with a force acting in a plane, it's important to understand how to break it down into its components. This means expressing the force in terms of its horizontal (x) and vertical (y) parts. In physics, we use trigonometry for this.
  • The horizontal component, or the x-component, of a force can be found using the equation: \( F_x = F \cos(\theta) \).
  • The vertical component, or the y-component, is calculated with: \( F_y = F \sin(\theta) \).
Here's a simple way to remember this:
  • Cosine is associated with the horizontal direction, aligning with the x-axis.
  • Sine is associated with the vertical direction, aligning with the y-axis.
By calculating these components, you are able to work with forces in a two-dimensional plane effectively.
Dot Product
The dot product is a crucial concept when calculating work done by a force. It involves two vectors and provides a scalar (single number) result. To compute the work done by a force acting on an object over a displacement, you multiply the force vector by the displacement vector using the dot product.
  • If \( \vec{F} = F_x \ \hat{i} + F_y \ \hat{j} \) is the force vector and \( \vec{d} = d_x \ \hat{i} + d_y \ \hat{j} \) is the displacement vector, the dot product equation is:
\[ W = \vec{F} \cdot \vec{d} = F_x \cdot d_x + F_y \cdot d_y \]The beauty of the dot product is that it considers only the amount of force used in the direction of movement. Only forces aligned with displacement contribute to work. This concept is key to understand energy transfer through forces.
Displacement Vector
A displacement vector provides both the direction and distance over which an object moves. It's represented with components along the x and y axes in two-dimensional space.
  • For example, a displacement that shifts an object 2 meters right and 4 meters down is written as \( \vec{d} = 2 \hat{i} - 4 \hat{j} \).
  • In this form, \( \hat{i} \) and \( \hat{j} \) are unit vectors along the x and y axes, respectively.
  • This method of representation simplifies calculations, like those for work or when using the dot product.
The displacement vector is offered as a straightforward method to capture movement across a plane in a single expression.
Trigonometry in Physics
Trigonometry serves as the backbone for solving physics problems involving angles and forces. It simplifies the process of finding components of forces and various vectors.
  • Key trigonometric functions (sine, cosine, and tangent) relate the angles of a triangle to its sides, which is fundamental for component calculations.
  • For angles not commonly memorized, using a calculator helps determine the values of the trigonometric functions effortlessly.
  • Forces forming an angle with a reference direction (like the x-axis) require these calculations for practical solutions.
Having an understanding of trigonometry aids in breaking down complex problems and results in more accurate physical interpretations.

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Most popular questions from this chapter

(a) In 1975 the roof of Montreal's Velodrome, with a weight of \(360 \mathrm{kN}\), was lifted by \(10 \mathrm{~cm}\) so that it could be centered. How much work was done on the roof by the forces making the lift? (b) In 1960 a Tampa, Florida, mother reportedly raised one end of a car that had fallen onto her son when a jack failed. If her panic lift effectively raised \(4000 \mathrm{~N}\) (about \(\frac{1}{4}\) of the car's weight) by \(5.0 \mathrm{~cm}\), how much work did her force do on the car?

During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapults that are made with surgical hose mounted on a window frame. A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hooke's law with a spring constant of \(100 \mathrm{~N} / \mathrm{m}\). If the hose is stretched by \(5.00 \mathrm{~m}\) and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length?

A horse pulls a cart with a force of \(40 \mathrm{lb}\) at an angle of \(30^{\circ}\) above the horizontal and moves along at a speed of \(6.0 \mathrm{mi} / \mathrm{h} .\) (a) How much work does the force do in \(10 \mathrm{~min} ?\) (b) What is the average power (in horsepower) of the force?

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of \(10.0\) m: (a) the initially stationary spelunker is accelerated to a speed of \(5.00 \mathrm{~m} / \mathrm{s} ;(\mathrm{b})\) he is then lifted at the constant speed of \(5.00 \mathrm{~m} / \mathrm{s} ;\) (c) finally he is decelerated to zero speed. How much work is done on the \(80.0 \mathrm{~kg}\) rescuee by the force lifting him during each stage?

An iceboat is at rest on a frictionless frozen lake when a sudden wind exerts a constant force of \(200 \mathrm{~N}\), toward the east, on the boat. Due to the angle of the sail, the wind causes the boat to slide in a straight line for a distance of \(8.0 \mathrm{~m}\) in a direction \(20^{\circ}\) north of east. What is the kinetic energy of the iceboat at the end of that \(8.0 \mathrm{~m} ?\)
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