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Chapter 6: Problem 3

A factory worker pushes a \(30.0-\mathrm{kg}\) 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
(a) 73.6 N; (b) 331.1 J; (c) -331.1 J; (d) 0 J for both normal force and gravity; (e) 0 J total work done.

Step by step solution

01

Determine the Force of Friction

The force of kinetic friction (\(f_k\)) can be calculated using the formula \(f_k = \mu_k \times N\), where \(\mu_k = 0.25\) is the coefficient of kinetic friction, and \(N\) is the normal force. For a level floor, the normal force \(N\) equals the weight of the crate, which is \(mg\) (mass times gravity). Thus, \(N = 30.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 294.3 \, \text{N}\). Hence, \(f_k = 0.25 \times 294.3 \, \text{N} = 73.575 \, \text{N}\).
02

Calculate the Force Required by the Worker

The worker needs to apply a force equal in magnitude to the force of friction to maintain constant velocity (i.e., no net acceleration). Therefore, the required force \(F\) by the worker is 73.575 N.
03

Calculate the Work Done by the Worker

Work done is given by the formula \( W = F \times d \times \cos(\theta) \), where \(\theta\) is the angle between the force direction and the direction of motion. Here, \(\theta = 0^\circ\), so \(\cos(0^\circ) = 1\). Thus, \( W = 73.575 \, \text{N} \times 4.5 \, \text{m} = 331.0875 \, \text{J}\).
04

Calculate the Work Done by Friction

Friction also opposes the motion with the same magnitude, hence, the work done by friction is \( W_f = -f_k \times d\). Substituting the values, \( W_f = -73.575 \, \text{N} \times 4.5 \, \text{m} = -331.0875 \, \text{J}\).
05

Calculate the Work Done by Normal Force and Gravity

The normal force and gravitational force act perpendicular to the horizontal motion; hence, the work done by these forces is zero since \( W = F \times d \times \cos(\theta) \) and \(\cos(90^\circ) = 0\).
06

Calculate the Total Work Done on the Crate

The total work done on the crate is the sum of work done by all forces: \(331.0875 \, \text{J} + (-331.0875 \, \text{J}) + 0 + 0 = 0 \, \text{J}\).

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Key Concepts

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

Kinetic Friction
Kinetic friction is a force that opposes the motion of two surfaces sliding past each other. In our scenario, this force acts between the crate and the floor. It depends on two main factors:
  • The coefficient of kinetic friction (\(\mu_k\)) - a value that represents how two surfaces interact;
  • The normal force (\(N\)), which is the force perpendicular to the surfaces in contact.
The force of kinetic friction (\(f_k\)) can be calculated using:\[ f_k = \mu_k \times N \]For our example, \(\mu_k\) is given as 0.25. Knowing that the normal force equals the gravity force on the crate, we plug in the numbers to find \(f_k = 0.25 \times 294.3 \text{ N} = 73.575 \text{ N}\). This frictional force is what the worker must overcome to move the crate.
Normal Force
Normal force is a force that acts perpendicular to the surface an object is resting on. For the crate, this force is a direct result of gravity pulling it down towards the floor. Conveniently, on a level surface where horizontal motion is our focus, the normal force (\( N \)) is equal to the gravitational force exerted on the crate. The gravitational force can be expressed as:\[ N = mg \]Where \( m \) is the mass (30 kg) and \( g \) is the acceleration due to gravity (\( 9.81 \text{ m/s}^2 \)). This results in a normal force of \( 294.3 \text{ N} \). Since normal force acts perpendicular to motion in this case, it does no work.
Constant Velocity
When an object moves at a constant velocity, it means its speed and direction remain unchanged. This only occurs when all the forces balancing the object are equal, resulting in no net force.For the worker pushing the crate:
  • The worker's force equals the kinetic friction force.
  • Both forces are equal: \( 73.575 \text{ N} \).
This balance keeps the crate moving at a steady rate and ensures no acceleration is occurring. Constant velocity indicates that the forces in the direction of motion are perfectly balanced.
Work Done by Forces
Work is defined as a force applied over a distance, typically expressed as:\[ W = F \times d \times \cos(\theta) \]Here, \( F \) is the force, \( d \) is the distance, and \( \theta \) is the angle between the force and the direction of motion.For the worker:
  • The work done equals: \( 73.575 \text{ N} \times 4.5 \text{ m} \times 1 \) (since the force is directly along the direction of motion) = \( 331.0875 \text{ J} \).
Similarly, the work done by kinetic friction is negative, as it works against movement:
  • This results in \( -331.0875 \text{ J} \) of work by friction.
Normal force and gravity do no work here since they act perpendicular to the distance moved, resulting in zero work from these forces due to \( \cos(90^\circ) = 0 \). As a result, the total work done is zero, illustrating energy balance in constant velocity motion.

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Most popular questions from this chapter

Magnetar. On December \(27,2004\) , astronomers observed the greatest flash of light ever recorded from outside the solar system. It came from the highly magnetic neutron star SGR \(1806-20\) (a magnetar). During 0.20 \(\mathrm{s}\) , this star released as much energy as our sun does in \(250,000\) years. If \(P\) is the average power output of our sun, what was the average power output (in terms of \(P\)) of this magnetar?

How many joules of energy does a 100 -watt light bulb use per hour? How fast would a \(70-\mathrm{kg}\) person have to run to have that amount of kinetic energy?

A soccer ball with mass 0.420 \(\mathrm{kg}\) is initially moving with speed 2.00 \(\mathrm{m} / \mathrm{s}\) . A soccer player kicks the ball, exerting a constant force of magnitude 40.0 \(\mathrm{N}\) in the same direction as the ball's motion. Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 \(\mathrm{m} / \mathrm{s} ?\)

A \(4.00-\mathrm{kg}\) block of ice is placed against a horizontal spring that has force constant \(k=200 \mathrm{N} / \mathrm{m}\) and is compressed 0.025 \(\mathrm{m}\) . The spring is released and accelerates the block along a horizontal surface. You can ignore friction and the mass of the spring. (a) Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its nncompressed length. (b) What is the speed of the block after it leaves the spring?

The spring of a spring gun has force constant \(k=400 \mathrm{N} / \mathrm{m}\) and negligible mass. The spring is compressed \(6.00 \mathrm{cm},\) and a ball with mass 0.0300 \(\mathrm{kg}\) is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 \(\mathrm{cm}\) long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal. (a) Calculate the speed with which the ball leaves the barrel if you can ignore friction. (b) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 \(\mathrm{N}\) acts on the ball as it moves along the barrel. (c) For the situation in part (b), at what position along the barrel does the ball have the greatest speed, and what is that speed? (In this case, the maximum speed does not occur at the end of the barrel)
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