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Chapter 6: Problem 3

A factory worker pushes a 30.0 -kg crate a distance of 4.5 \(\mathrm{m}\) along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25 . (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?

Short Answer

Expert verified
The worker applies 73.5 N; he does 330.75 J of work; friction does -330.75 J; normal force and gravity do 0 J; total work is 0 J.

Step by step solution

01

Calculate the Force Required to Move the Crate

To keep the crate moving at a constant velocity, the worker needs to apply a force that balances the force of kinetic friction. The force of friction \(F_f\) is given by \(F_f = \mu_k \cdot N\), where \(\mu_k\) is the coefficient of kinetic friction and \(N\) is the normal force. On a level surface, the normal force \(N\) is equal to the gravitational force on the crate, which is \(N = m \cdot g = 30.0 \cdot 9.8 = 294.0 \; N\). Thus, \(F_f = 0.25 \times 294.0 = 73.5 \, N\).
02

Calculate the Work Done by the Worker

Work is calculated as \(W = F \cdot d\), where \(F\) is the force applied and \(d\) is the distance over which the force is applied. Since the crate moves at constant velocity, the worker's force is equal to the force of friction, \(73.5 \, N\). The distance moved is \(4.5 \, m\), so the work done by the worker is \(W = 73.5 \times 4.5 = 330.75 \, J\).
03

Calculate the Work Done by Friction

The work done by friction is also \(W = F \cdot d\) but, since friction acts opposite to the direction of movement, this work is negative. Therefore, \(W = -73.5 \times 4.5 = -330.75 \, J\).
04

Calculate the Work Done by the Normal Force

The normal force acts perpendicular to the movement of the crate. Since work is the dot product of force and displacement and there is no displacement in the direction of the normal force, the work done by the normal force is \(0 \, J\).
05

Calculate the Work Done by Gravity

Similar to the normal force, gravity acts vertically while the displacement is horizontal. Therefore, the work done by gravity is \(0 \, J\).
06

Calculate the Total Work Done on the Crate

The total work done on the crate is the algebraic sum of the work done by all forces: \(330.75 \, J\) by the worker, \(-330.75 \, J\) by friction, \(0 \, J\) by the normal force, and \(0 \, J\) by gravity. Thus, the total work done is \(330.75 + (-330.75) + 0 + 0 = 0 \, J\).

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Key Concepts

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

Kinetic friction
Kinetic friction occurs when two objects slide against each other. It is a force that resists the relative motion of the surfaces in contact. In this scenario, the crate sliding across the factory floor experiences kinetic friction from the floor. This frictional force is calculated by multiplying the coefficient of kinetic friction (\(\mu_k\)) by the normal force (\(N\)).
  • Coefficient of kinetic friction, \(\mu_k\), is a value that depends on the two surfaces in contact. In our problem, \(\mu_k = 0.25\).
  • The normal force \(N\) is the force perpendicular to the surfaces in contact. On a level surface, it equals the weight of the object, calculated as \(m \cdot g\) where \(m\) is the mass and \(g\) is the acceleration due to gravity (approx. \(9.8 \, \text{m/s}^2\)).
Thus, for the worker pushing the crate, the force of kinetic friction opposing the motion is \(73.5 \, N\). This force must be balanced by the worker's push to keep the crate moving at constant speed.
Work-energy principle
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. Work is calculated by the product of the force applied and the distance over which the force is applied in the direction of the force: \(W = F \cdot d\).
Understanding Work Done by Different Forces:
  • Work by the Worker: To move the crate at constant velocity, the work done by the worker equals the work done against kinetic friction. In our case, this is \(330.75 \, J\).
  • Work by Friction: Kinetic friction does negative work, meaning it takes energy from the system. It is also \(-330.75 \, J\) in this instance, since friction acts in the opposite direction of movement.
  • Total Work: Total work accounts for all forces: worker's force, friction, normal force, and gravitational force. Here, the algebraic sum results in \(0 \, J\), hence no net change in kinetic energy, which aligns with the constant velocity.
Newton's laws of motion
Newton's laws provide the foundation for analyzing the movements and interactions of objects. Particularly relevant here is Newton's first law, often related to inertia, and Newton's second law regarding force and acceleration.
  • First Law (Inertia): An object at rest remains at rest, and an object in motion remains in motion at a constant velocity unless acted upon by a net external force. In this problem, the crate moves at a constant velocity because the net external force is zero. The force the worker applies perfectly balances the frictional force.
  • Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass: \(F_{net} = m \cdot a\). For our crate, \(a = 0\) because it moves uniformly; hence \(F_{net} = 0\).
The worker's force and friction maintain a neutral balance allowing the crate to glide smoothly across the floor.
Forces in physics
Forces are fundamental to understanding how objects interact and move. In physics, forces can cause objects to start moving, speed up, slow down, or change direction. Different types of forces come into play in various situations, such as the crate's movement.
Key Forces in This Scenario:
  • Applied Force: This is the force exerted by the factory worker on the crate. It needs to be equal and opposite to the force of friction to allow movement at a constant velocity.
  • Frictional Force: Acts opposite to the direction of motion, calculated with the coefficient of kinetic friction and normal force.
  • Normal Force: Acts perpendicular to the surface, equal to the gravitational force when the surface is flat.
  • Gravitational Force: Pulls objects toward the center of the Earth, impacting the normal force but doing no work in the horizontal motion.
Understanding these forces helps clarify how the worker moves the crate without changing its velocity. The balance of forces assures steady movement on the factory floor.

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Most popular questions from this chapter

Some Typical Kinetic Energies. (a) In the Bohr model of the atom, the ground- state electron in hydrogen has an orbital speed of 2190 \(\mathrm{km} / \mathrm{s} .\) What is its kinetic energy? (Consult Appendix F.) (b) If you drop a 1.0-kg weight (about 2 lb) from a height of 1.0 \(\mathrm{m}\) , how many joules of kinetic energy will it have when it reaches the ground? (c) Is it reasonable that a \(30-\mathrm{kg}\) child could run fast enough to have 100 \(\mathrm{J}\) of kinetic energy?

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 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 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?

CALC A force in the \(+x\) -direction with magnitude \(F(x)=18.0 \mathrm{N}-(0.530 \mathrm{N} / \mathrm{m}) x\) is applied to a 6.00 -kg box that is sitting on the horizontal, frictionless surface of a frozen lake. \(F(x)\) is the only horizontal force on the box. If the box is initially at rest at \(x=0,\) what is its speed after it has traveled 14.0 \(\mathrm{m}\) ?

CALC An airplane in flight is subject to an air resistance force proportional to the square of its speed \(v .\) But there is an additional resistive force because the airplane has wings. Air flowing over the wings is pushed down and slightly forward, so from Newton's third law the air exerts a force on the wings and airplane that is up and slightly backward (Fig. P6.104). The upward force is the lift force that keeps the airplane aloft, and the backward force is called induced drag. At flying speeds, induced drag is inversely proportional to \(v^{2},\) so that the total air resistance force can be expressed by \(F_{\text { air }}=\alpha v^{2}+\beta / v^{2},\) where \(\alpha\) and \(\beta\) are positive constants that depend on the shape and size of the airplane and the density of the air. For a Cessna \(150,\) a small single-engine airplane, \(\alpha=0.30 \mathrm{N} \cdot \mathrm{s}^{2} / \mathrm{m}^{2}\) and \(\beta=3.5 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} .\) In steady flight, the engine must provide a forward force that exactly balances the air resistance force. (a) Calculate the speed (in \(\mathrm{km} / \mathrm{h} )\) at which this airplane will have the maximum range (that is, travel the greatest distance) for a given quantity of fuel. (b) Calculate the speed (in \(\mathrm{km} / \mathrm{h} )\) for which the airplane will have the maximum endurance (that is, remain in the air the longest time).

A physics professor is pushed up a ramp inclined upward at \(30.0^{\circ}\) above the horizontal al as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and chair is 85.0 \(\mathrm{kg} .\) He is pushed 2.50 \(\mathrm{m}\) along the incline by a group of students who together exert a constant horizontal force of 600 \(\mathrm{N} .\) The professor's speed at the bottom of the ramp is 2.00 \(\mathrm{m} / \mathrm{s} .\) Use the work-energy theorem to find his speed at the top of the ramp.
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