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Chapter 6: Problem 7

A person pushes a \(16.0\) -kg shopping cart at a constant velocity for a distance of \(22.0 \mathrm{~m}\). She pushes in a direction \(29.0^{\circ}\) below the horizontal. A \(48.0\) - \(\mathrm{N}\) frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.

Short Answer

Expert verified
The shopper exerts approximately 54.9 N. The work done by the pushing force is approximately 1054.1 J, by friction is -1056 J, and by gravity is 0 J.

Step by step solution

01

Determine Shoppers Force in Horizontal

Since the shopping cart is pushed at a constant velocity, the net horizontal force is zero. This means the horizontal component of the pushing force must balance the frictional force. The horizontal component of the force can be represented as:\[F_{x} = F \cdot \cos(29^{\circ})\]Where \(F\) is the force exerted by the shopper. This component should equal the frictional force of \(48.0\, \mathrm{N}\), i.e.\[F \cdot \cos(29^{\circ}) = 48.0 \mathrm{~N}\]Now, solve for \(F\):\[F = \frac{48.0\, \mathrm{N}}{\cos(29^{\circ})}\]\[\approx 54.9\, \mathrm{N}\]
02

Calculate the Work Done by the Pushing Force

The work done by the force is the product of the component of the force in the direction of movement and the displacement. We calculate it as:\[W_{push} = F \cdot d \cdot \cos(\theta)\]Substitute the values:\[W_{push} = 54.9\, \mathrm{N} \cdot 22.0\, \mathrm{m} \cdot \cos(29^{\circ})\]Calculate:\[W_{push} \approx 54.9 \cdot 22.0 \cdot 0.8746 \approx 1054.1\, \mathrm{J}\]
03

Calculate the Work Done by the Frictional Force

The work done by the frictional force is negative because it opposes the motion. It can be calculated as:\[W_{friction} = -f \cdot d\]Where \(f = 48.0\, \mathrm{N}\) and \(d=22.0\, \mathrm{m}\). So:\[W_{friction} = -(48.0 \times 22.0)\]Calculate:\[W_{friction} = -1056\, \mathrm{J}\]
04

Calculate the Work Done by Gravitational Force

The gravitational force does no work in this scenario since there's no vertical displacement of the cart (it's moving on a horizontal surface). The work done by gravity is:\[W_{gravity} = 0\, \mathrm{J}\]

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Key Concepts

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

Work and Energy
In the realm of physics, understanding work and energy is a fundamental concept. Work is done when a force causes an object to move in the direction of the force. To calculate work done, you can use the formula:
  • Work, \(W = F imes d imes ext{cos}(\theta)\)
  • \( F \) is the force applied, \( d \) is the displacement, and \( \theta \) is the angle between the force direction and the displacement direction.
In our scenario, the shopper pushes the cart over a distance of \( 22.0 \, \mathrm{m} \). Since the cart is pushed at an angle \( 29.0^{\circ} \) below the horizontal, only the horizontal component of her push does work. It鈥檚 important to use cosine of the angle, as it relates the force in the direction of motion. Remember, energy in physics, especially mechanical energy, is associated with the ability to do work.
Constant Velocity
A constant velocity implies that an object moves with uniform speed in a straight line. It is significant here because it means that the net force acting on the shopping cart is zero. When the velocity is constant, any forces that cause motion are perfectly balanced by opposing forces, like friction in this case.In simpler terms:
  • The force the shopper applies horizontally exactly counteracts the frictional force.
  • This ensures that the shopping cart does not accelerate but moves at a steady pace.
For our exercise, the frictional force is \( 48.0 \, \mathrm{N} \). Hence, the horizontal force component applied by the shopper must also be \( 48.0 \, \mathrm{N} \), ensuring the cart鈥檚 constant velocity.
Force Components
Understanding force components is crucial, especially when dealing with angles. Any force can be broken down into two components: horizontal and vertical.For a force \( F \) applied at an angle \( \theta \) below the horizontal:
  • The horizontal component is calculated as \( F_x = F \, \text{cos}(\theta) \).
  • The vertical component is \( F_y = F \, \text{sin}(\theta) \).
In our problem, since the force is applied \( 29.0^{\circ} \) below the horizontal, we use the cosine of this angle to find the horizontal component. This horizontal component is what counteracts the frictional force, leading to no net force and constant velocity. Hence, force components help us understand which part of the force does what, be it moving the cart or negotiating friction.
Frictional Force
Frictional force is a resistive force that opposes the motion or attempted motion of an object over a surface. In physics, friction is often considered a necessary evil, as it enables us to walk or drive, but it also opposes motion, requiring energy to overcome.Key points on frictional force:
  • It acts in the opposite direction to the movement.
  • In our exercise, the frictional force is quantified as \( 48.0 \, \mathrm{N} \).
  • This force must be overcome to maintain the cart's motion at constant velocity.
Since no vertical displacement occurs, the frictional force strictly opposes horizontal motion. This opposition is why the work done by friction is negative, representing energy lost from the system due to friction.

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Most popular questions from this chapter

A \(5.00 \times 10^{2}-\) kg hot-air balloon takes off from rest at the surface of the earth. The nonconservative wind and lift forces take the balloon up, doing \(+9.70 \times 10^{4} \mathrm{~J}\) of work on the balloon in the process. At what height above the surface of the earth does the balloon have a speed of \(8.00 \mathrm{~m} / \mathrm{s} ?\)

Two cars, \(A\) and \(B\), are traveling with the same speed of \(40.0 \mathrm{~m} / \mathrm{s}\), each having started from rest. Car A has a mass of \(1.20 \times 10^{3} \mathrm{~kg}\), and car \(\mathrm{B}\) has a mass of \(2.00 \times 10^{3} \mathrm{~kg} .\) Compared to the work required to bring car A up to speed, how much additional work is required to bring car B up to speed?

Relative to the ground, what is the gravitational potential energy of a 55.0 -kg person who is at the top of the Sears Tower, a height of \(443 \mathrm{~m}\) above the ground?

A \(2.00\) -kg rock is released from rest at a height of \(20.0 \mathrm{~m}\). Ignore air resistance and determine the kinetic energy, gravitational potential energy, and total mechanical energy at each of the following heights: \(20.0,10.0\), and \(0 \mathrm{~m}\).

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 \(0.70 \mathrm{~m} / \mathrm{s}^{2}\) and is lifted from rest through a distance of \(11 \mathrm{~m}\). (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.
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