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Chapter 6: Problem 8

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force \(\vec{F}=(30 \mathrm{N}) \hat{t}-(40 \mathrm{N}) \hat{\jmath}\) to the cart as it undergoes a displacement \(\vec{s}=(-9.0 \mathrm{m}) \hat{\imath}-(3.0 \mathrm{m}) \hat{\mathrm{j}}\) How much work does the force you apply do on the grocery cart?

Short Answer

Expert verified
The work done on the grocery cart is \(-150 \, \text{J}\).

Step by step solution

01

Understanding Work Done by a Force

The work done by a force can be calculated using the dot product of the force vector \( \vec{F} \) and the displacement vector \( \vec{s} \). The formula for work is given by: \[ W = \vec{F} \cdot \vec{s} = F_x s_x + F_y s_y \]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

Identify Components of Force and Displacement

Identify the components of the vectors based on their unit vector notation:- \( \vec{F} = (30 \, \text{N}) \hat{\imath} - (40 \, \text{N}) \hat{\jmath} \) gives us \( F_x = 30 \, \text{N} \) and \( F_y = -40 \, \text{N} \).- \( \vec{s} = (-9.0 \, \text{m}) \hat{\imath} - (3.0 \, \text{m}) \hat{\jmath} \) gives us \( s_x = -9.0 \, \text{m} \) and \( s_y = -3.0 \, \text{m} \).
03

Calculate Each Component of Work

Plug the components into the work formula:\[ W_x = F_x \times s_x = 30 \, \text{N} \times (-9.0 \, \text{m}) = -270 \, \text{J} \]\[ W_y = F_y \times s_y = (-40 \, \text{N}) \times (-3.0 \, \text{m}) = 120 \, \text{J} \]
04

Sum the Work Components

Add the calculated work components together to find the total work done:\[ W = W_x + W_y = -270 \, \text{J} + 120 \, \text{J} = -150 \, \text{J} \]
05

Interpret the Result

The total work done by the force on the grocery cart is \(-150 \, \text{J}\). The negative sign indicates that the direction of the force is opposite to that of the displacement.

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Key Concepts

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

Dot Product
The dot product is a way to multiply two vectors, resulting in a scalar quantity, not another vector. It's crucial when calculating work done by a force. You compute it by taking the product of the magnitudes of the two vectors and the cosine of the angle between them. However, when vectors are in component form, you can simplify this to a component-by-component multiplication:
  • Multiply corresponding components of the vectors.
  • Add these products together to get the dot product.
In our exercise, the dot product helps link the two separate quantities, the force vector and the displacement vector, to find the work done. When the angle between two vectors is 90 degrees, their dot product is zero, which implies no work is done if the force is perpendicular to the displacement.
Force Vector
A force vector represents a force applied in a particular direction. Forces are not just defined by their magnitude but also by their direction, which is why vectors are ideal to represent them. In the given problem, we have the force vector denoted as \( \vec{F} = (30 \, \text{N}) \hat{\imath} - (40 \, \text{N}) \hat{\jmath} \). This indicates:
  • A force of 30 Newtons in the positive x-direction (\( \hat{\imath} \)).
  • A force of 40 Newtons in the negative y-direction (\( \hat{\jmath} \)).
Understanding force vectors as components is essential since they help in calculating the effect of each part of the force separately, making it easier to compute total work done by a force acting at angles.
Displacement Vector
A displacement vector measures the change in position of an object and is a key aspect in understanding work done by a force. Unlike just distance, displacement is directional and can be represented with vectors. For the grocery cart, we have the displacement vector \( \vec{s} = (-9.0 \, \text{m}) \hat{\imath} - (3.0 \, \text{m}) \hat{\jmath} \):
  • The cart moves 9 meters in the negative x-direction.
  • It also shifts 3 meters in the negative y-direction.
This is different from the path taken, emphasizing direction from the start to the endpoint. In relation to work, it shows exactly how the object's position changes due to applied forces over a specific path, which must be considered along with force directions.
Components of Vectors
Vectors are often broken down into components to simplify calculations involving multiple directions. The components correspond to how much influence a vector has in each axis鈥攃ommonly x, y, and sometimes z.In resolving vectors into components:
  • Each vector with directions specified can be written as: \( \vec{V} = V_x \hat{\imath} + V_y \hat{\jmath} \) (in two dimensions).
  • If in three dimensions, add \( V_z \hat{k} \).
In the exercise, the force and displacement vectors are broken into x and y components:\( F_x = 30 \, \text{N} \), \( F_y = -40 \, \text{N} \), \( s_x = -9.0 \, \text{m} \), \( s_y = -3.0 \, \text{m} \).These components are key in the calculations,enabling the use of the dot product formula to compute work done, by simply multiplying corresponding components and summing their products.

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Most popular questions from this chapter

A sled with mass 8.00 \(\mathrm{kg}\) moves in a straight line on a frictionless horizontal surface. At one point in its path, its speed is 4.00 \(\mathrm{m} / \mathrm{s}\) : after it has traveled 2.50 \(\mathrm{m}\) beyond this point, its speed is 6.00 \(\mathrm{m} / \mathrm{s}\) . Use the work-energy theorem to find the force acting on the sled, assuming that this force is constant and that it acts in the direction of the sled's motion.

You and your bicycle have combined mass 80.0 \(\mathrm{kg}\) . When you reach the base of a bridge, you are traveling along the road at 5.00 \(\mathrm{m} / \mathrm{s}(\mathrm{Fig} .6 .35) .\) At the top of the bridge, you have climbed a vertical distance of 5.20 \(\mathrm{m}\) and have slowed to 1.50 \(\mathrm{m} / \mathrm{s}\) . You can ignore work done by friction and any inefficiency in the bike or your legs. (a) What is the total work done on you and your bicycle when you go from the base to the top of the bridge? (b) How much work have you done with the force you apply to the pedals?

Working Like a Horse. Your job is to lift \(30-k g\) crates a vertical distance of 0.90 \(\mathrm{m}\) from the ground onto the bed of a truck. (a) How many crates would you have to load onto the truck in 1 minute for the average power output you use to lift the crates to equal 0.50 \(\mathrm{hp} ?\) (b) How many crates for an average power output of 100 \(\mathrm{W} ?\)

An object is attracted toward the origin with a force given by \(F_{x}=-k / x^{2} .\) (Gravitational and electrical forces have this distance dependence.) (a) Calculate the work done by the force \(F_{x}\) when the object moves in the \(x\) -direction from \(x_{1}\) to \(x_{2}\) . If \(x_{2}>x_{1}\) , is the work done by \(F_{x}\) positive or negative? (b) The only other force acting on the object is a force that you exert with your hand to move the object slowly from \(x_{1}\) to \(x_{2}\) . How much work do you do? If \(x_{2}>x_{1},\) is the work you do positive or negative? (c) Explain the similarities and differences between your answers to parts (a) and (b).

Half of a Spring. (a) Suppose you cut a massless ideal spring in half. If the full spring had a force constant \(k\) , what is the force constant of each half, in terms of \(k ?\) (Hint: Think of the original spring as two equal halves, each producing the same force as the entire spring. Do you see why the forces must be equal? (b) If you cut the spring into three equal segments instead, what is the force constant of each one, in terms of \(k ?\)
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