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You push your physics book \(1.50 \mathrm{~m}\) along a horizontal tabletop with a horizontal push of \(2.40 \mathrm{~N}\) while the opposing force of friction is \(0.600 \mathrm{~N}\). How much work does each of the following forces do on the book: (a) your \(2.40 \mathrm{~N}\) push, (b) the friction force, (c) the normal force from the tabletop, and (d) gravity? (e) What is the net work done on the book?

Short Answer

Expert verified
The work done by the push is \(3.60 \mathrm{J}\), by the friction is \(-0.90 \mathrm{J}\), by the normal force is \(0 \mathrm{J}\), by the gravity is \(0 \mathrm{J}\), and the net work done on the book is \(2.70 \mathrm{J}\).

Step by step solution

01

Calculate the work done by your push

The pushing force and displacement are in the same direction, hence \(\theta = 0\). Substituting the values into the work formula, we get: \( W = F \cdot d \cdot \cos(\theta) = 2.40 \mathrm{N} \cdot 1.50 \mathrm{m} \cdot \cos(0) = 3.60 \mathrm{J}\)
02

Calculate the work done by the friction

The frictional force is opposite to the displacement, hence \(\theta = 180\degree\). Substituting the values into the work formula, we get: \( W = -0.600 \mathrm{N} \cdot 1.50 \mathrm{m} \cdot \cos(180\degree) = -0.90 \mathrm{J}\)
03

Calculate the work done by the normal force

The normal force is perpendicular to the direction of displacement, hence \(\theta = 90\degree\). Since the directional component of the force (\(\cos(\theta)\)) will be 0, the work done by the normal force will be 0.
04

Calculate the work done by the force of gravity

Gravity acts vertically downwards and the displacement is horizontal, thus these two vectors are perpendicular. We thus have \(\theta = 90\degree\). So like the normal force, the work done by gravity will also be 0.
05

Calculate the net work done on the book

The net work done on the book is the sum of the work done by all the forces, which is \(3.60 \mathrm{J} + -0.90 \mathrm{J} + 0 + 0 = 2.70 \mathrm{J}\)

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Key Concepts

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

Force and Motion
When we talk about force and motion, we are discussing the interaction that causes something to change speed, direction, or position. In this scenario, you are applying a force to move your physics book across the table. The push you give is a force that initiates motion. This force is measured in Newtons (N), and in this case, you apply a force of 2.40 N along a 1.50 m path.

When you apply a force, it can cause the object to accelerate. The relation between force, mass, and acceleration is given by Newton's Second Law: \[ F = ma \]
Where,
  • \( F \) is the force applied,
  • \( m \) is the mass of the object,
  • and \( a \) is the acceleration.

In the exercise, the force is responsible for moving the book across the table and overcoming any resistance like friction. Understanding this interaction is critical for solving problems involving motion.
Friction
Friction is the resistive force that opposes the motion or attempted motion of an object across a surface. It's what you felt as you tried to slide a book across a table and noticed resistance.

The force of friction acts in the opposite direction to the applied force. In this exercise, the frictional force was 0.600 N. This is why it acts against your push of 2.40 N.

Friction can be calculated by: \[ f_r = ext{friction coefficient} imes ext{normal force} \] The friction you encounter is essential since it prevents your book from gliding indefinitely without horizontal force applied.

Without friction, every push would be like moving stuff on an extremely smooth ice surface. It is key to almost every mechanical system in everyday life.
Normal Force
The normal force is the support force exerted by a surface to support the weight of an object resting on it. It acts perpendicular to the surface, and its main role is to balance out the gravitational pull on the object. Think of it as the invisible hand holding your book against the push-down force of gravity.

In the scenario, as the book rests on the table, the table exerts an upward normal force equal in magnitude and opposite in direction to the weight of the book. This keeps the book from crashing through the table. Since the force is perpendicular to the book's motion, it does not do any work.

It's important to understand that without the normal force, the book would simply sink into the table due to gravity. The normal force cancels out the vertical component of forces acting on an object on a flat surface, ensuring it remains in equilibrium.
Gravity
Gravity is the force that pulls objects towards each other, attracting them downwards towards the center of the Earth. It acts vertically downward. In our exercise, gravity is attempting to pull the book towards the floor while the normal force from the table keeps it at rest.

Gravity does work on objects moving up or down, but for horizontal motion like sliding a book across a table, no work is done by gravity. Work done by gravity is determined by: \[ W = mgh \] for vertical displacement, where
  • \( m \) is the mass,
  • \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \)),
  • and \( h \) is the height.

In this flat-motion case, the ideal gravitational work is zero, illustrating how different forces interact in the real world to create balanced systems.

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Most popular questions from this chapter

To stretch an ideal spring \(3.00 \mathrm{~cm}\) from its unstretched length, \(12.0 \mathrm{~J}\) of work must be done. (a) What is the force constant of this spring? (b) What magnitude force is needed to stretch the spring 3.00 \(\mathrm{cm}\) from its unstretched length? (c) How much work must be done to compress this spring \(4.00 \mathrm{~cm}\) from its unstretched length, and what force is needed to compress it this distance?

A block of mass \(m\) is released from rest at the top of an incline that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction between the block and incline is \(\mu_{k}\). The top of the incline is a vertical distance \(h\) above the bottom of the incline. Derive an expression for the work \(W_{f}\) done on the block by friction as it travels from the top of the incline to the bottom. When \(\alpha\) is decreased, does the magnitude of \(W_{f}\) increase or decrease?

A pump is required to lift \(800 \mathrm{~kg}\) of water (about 210 gallons) per minute from a well \(14.0 \mathrm{~m}\) deep and eject it with a speed of \(18.0 \mathrm{~m} / \mathrm{s}\). (a) How much work is done per minute in lifting the water? (b) How much work is done in giving the water the kinetic energy it has when ejected? (c) What must be the power output of the pump?

Leg Presses. As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff ideal springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do \(80.0 \mathrm{~J}\) of work when you compress the springs \(0.200 \mathrm{~m}\) from their uncompressed length. (a) What magnitude of force must you apply to hold the platform in this position? (b) How much additional work must you do to move the platform \(0.200 \mathrm{~m}\) farther, and what maximum force must you apply?

The upper end of a light rope of length \(L=0.600 \mathrm{~m}\) is attached to the ceiling, and a small steel ball with mass \(m=0.200 \mathrm{~kg}\) is suspended from the lower end of the rope. Initially the ball is at rest and the rope is vertical. Then a force \(\vec{F}\) with constant magnitude \(F=0.760 \mathrm{~N}\) and a direction that is maintained tangential to the path of the ball is applied and the ball moves in an arc of a circle of radius \(L\). What is the speed of the ball when the rope makes an angle \(\alpha=37.0^{\circ}\) with the vertical?

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