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Chapter 9: Problem 26

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

Short Answer

Expert verified
The work done is \( -44.64 \text{ J} \).

Step by step solution

01

Understand the work formula

Work done by a force is given by the formula \( W = \textbf{F} \bullet \textbf{d} \), where \( \textbf{F} \) is the force vector and \( \textbf{d} \) is the displacement vector.
02

Resolve the Force Vector

Resolve the force vector \( \textbf{F} \) into its components. Given the magnitude is \(10\) N and the angle is \(150^{\text{\circ}}\), use the formulas \( F_x = F \times \text{cos}\theta \) and \( F_y = F \times \text{sin}\theta \) to get \( F_x = 10 \times \text{cos}(150^{\text{\circ}}) = -5 \text{ N} \) and \( F_y = 10 \times \text{sin}(150^{\text{\circ}}) = 8.66 \text{ N} \).
03

Displacement Vector

The displacement vector \( \textbf{d} \) is given by \( 2.0 \hat{i} - 4.0 \hat{j} \). So, its components are \( d_x = 2.0 \text{ m} \) and \( d_y = -4.0 \text{ m} \).
04

Apply the Dot Product

Calculate the dot product of \( \textbf{F} \) and \( \textbf{d} \). The dot product formula is given by \( \textbf{F} \bullet \textbf{d} = F_x \times d_x + F_y \times d_y \). Substituting the values, we get: \( (-5) \times 2.0 + 8.66 \times (-4.0) = -10 - 34.64 = -44.64 \).
05

Calculate Work Done

The total work done by the force is \( -44.64 \text{ J} \) (Joules). A negative value indicates that the force is doing work against the motion of the object.

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Key Concepts

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

Constant Force
In physics, a constant force is a force that remains unchanged over time. It has a fixed magnitude and direction. This makes calculations easier because we don't have to deal with varying values. It's like pushing a box across a room with the same force throughout. No matter where the box is, the force you apply remains the same.
Understanding constant forces helps make many physics problems more approachable. When given a constant force in an exercise, remember its properties stay the same during the object's movement.
Work Formula
The concept of work in physics is different from everyday use. In physics, work is done when a force causes an object to move. The formula to calculate work is given by:
\( W= \textbf{F} \bullet \textbf{d} \)
Where: \(\textbf{F}\) is the force vector and \(\textbf{d}\) is the displacement vector. Notice the dot product symbol (\( \bullet \)) here; it means this is a vector operation. The result of the dot product will give you a scalar value, which is the work done.
Always ensure you have both the force and displacement before applying this formula.
Vector Components
Vectors have both magnitude and direction. When we break a vector into its components, we find out how much of the vector acts in the horizontal (x-axis) and vertical (y-axis) directions. This is called resolving the vector.
For example, if we have a force vector with magnitude 10 N at an angle of 150 degrees, we'll use trigonometric functions to find its components:
  • \( F_x = F \times \text{cos}\theta \)
  • \( F_y = F \times \text{sin}\theta \)

This way, you get two new vectors \(F_x\) and \(F_y\) which are easier to work with individually.
Dot Product
The dot product is a way to multiply two vectors that result in a scalar quantity, not another vector. It's used to determine the work done by a force over a displacement. The formula for the dot product of two vectors is:
\( \textbf{F} \bullet \textbf{d} = F_x \times d_x + F_y \times d_y \)
Here, you multiply the x-components of both vectors together and the y-components together. Then add those results to get the final scalar value. It's helpful because it simplifies the interaction of vectors into a single number, making further calculations easier.
Displacement Vector
Displacement in physics means how far an object has moved from its original position. This is represented by a displacement vector. Unlike distance, displacement is directional. For example, if an object moves from the origin to the point (2.0 m, -4.0 m), the displacement vector is written as \( 2.0 \hat{\text{i}} - 4.0 \hat{\text{j}} \)
This tells us how far and in what direction (up or down and left or right) the object has moved. It's crucial in calculating work done because we need to know both the distance and direction of the movement.

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Most popular questions from this chapter

A Particle Moves A particle moves along a straight path through displacement \(\vec{d}=(8 \mathrm{~m}) \hat{\mathrm{i}}-(c \mathrm{~m}) \hat{\mathrm{j}}\) while force \(\vec{F}=\) \((2 \mathrm{~N}) \mathrm{i}-(4 \mathrm{~N}) \hat{\mathrm{j}}\) acts on it. (Other forces also act on the particle.) What is the value of \(c\) if the work done by \(\vec{F}\) on the particle is (a) zero, (b) positive, and (c) negative?

Two Pulleys and a Canister In Fig. \(9-28\), a cord runs around two massless, frictionless pulleys; a canister with mass \(m=20 \mathrm{~kg}\) hangs from one pulley; and you exert a force \(\vec{F}\) on the free end of the cord. (a) What must be the magnitude of \(\vec{F}\) if you are to lift the canister at a constant speed? (b) To lift the canister by \(2.0 \mathrm{~cm}\), how far must you pull the free end of the cord? During that lift, what is the work done on the canister by (c) your force (via the cord) and (d) the gravitational force on the canister? (Hint: When a cord loops around a pulley as shown, it pulls on the pulley with a net force that is twice the tension in the cord.)

Lowering a Block A cord is used to vertically lower an initially stationary block of mass \(M\) at a constant downward acceleration of \(g / 4\). When the block has fallen a distance \(d\), find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block.

A Proton is Accelerated A proton (mass \(m=1.67 \times 10^{-27} \mathrm{~kg}\) ) is being accelerated along a straight line at \(3.6 \times 10^{13} \mathrm{~m} / \mathrm{s}^{2}\) in a machine. If the proton has an initial speed of \(2.4 \times 10^{7} \mathrm{~m} / \mathrm{s}\) and travels \(3.5 \mathrm{~cm}\), what then is (a) its speed and (b) the increase in its kinetic energy?

Given \(x(t)\) A force acts on a \(3.0 \mathrm{~kg}\) particle-like object in such a way that the position of the object as a function of time is given by \(x=(3 \mathrm{~m} / \mathrm{s}) t-\left(4 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}+\left(1 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3}\) with \(x\) in meters and \(t\) in seconds Find the work done on the object by the force from \(t_{1}=0.0 \mathrm{~s}\) to \(t_{2}=4.0 \mathrm{~s}\). (Hint: What are the speeds at those times?)
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