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Bonnie and Clyde are sliding a \(300 \mathrm{kg}\) bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with \(385 \mathrm{N}\) of force while Bonnie pulls forward on a rope with \(350 \mathrm{N}\) of force. What is the safe's coefficient of kinetic friction on the bank floor?

Short Answer

Expert verified
The coefficient of kinetic friction between the safe and the bank floor is 0.25.

Step by step solution

01

Determine the Net Force

As the safe is moving with constant speed, the net force acting on it is zero. Considering Bonnie and Clyde are exerting forces in the same direction (Bonnie with a force of \(350 \mathrm{N}\) and Clyde with \(385 \mathrm{N}\)), the total force pushing the safe is \(385 \mathrm{N} + 350 \mathrm{N} = 735 \mathrm{N}\). The friction acts in the opposite direction with the same magnitude, so the friction force is \(735 \mathrm{N}\).
02

Determine the Normal Force

The normal force is the force exerted by a surface that supports the weight of an object resting on it. It acts perpendicular to the surface. For an object on a flat surface without vertical acceleration, the normal force equals the weight of the object. So, the normal force equal to the weight of the safe is \(300 \mathrm{kg} \times 9.8 \mathrm{m/s^2} = 2940 \mathrm{N}\).
03

Calculate the Coefficient of Kinetic Friction

The coefficient of kinetic friction can be calculated using the formula: \( \mu = \frac{F_f}{F_n} \), where \(F_f\) is the friction force and \(F_n\) is the normal force. Therefore, \(\mu = \frac{735 \mathrm{N}}{2940 \mathrm{N}} = 0.25\)

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Key Concepts

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

Friction Force
When two surfaces move against each other, friction comes into play. In this scenario, Bonnie and Clyde pushing and pulling the safe generates a friction force. This force opposes the motion of the safe, working against the direction of the applied forces from Bonnie and Clyde. The force of friction is responsible for slowing down or eventually stopping the movement if no continuous force is applied.

Friction force helps determine how much effort is needed to keep an object moving. For instance, in the exercise, Bonnie and Clyde need to balance their exerted force with the friction force. Since the safe moves at a constant speed, the friction force must be equal to the total applied force, ensuring there is no net force acting on the object. This is why the friction force is equal to 735 N, countering the combined force exerted by Bonnie and Clyde.

Understanding friction and how it acts will allow you to predict and calculate the needed effort when dealing with different surfaces or materials.
Normal Force
The concept of normal force is essential in understanding how objects interact with surfaces. It is the force that a surface exerts perpendicular to itself, counteracting the weight of an object resting on it. In simple terms, the normal force keeps objects from falling through the surface they sit on.

In the case of Bonnie and Clyde, the safe is sitting on a flat floor, meaning there's no vertical acceleration. Therefore, the normal force is simply the weight of the safe. You calculate it by multiplying the safe's mass by the gravitational acceleration, which gives us a normal force of 2940 N.

Knowing the normal force is crucial for calculating other forces such as friction, as it typically serves as a part of these calculations. This force plays a vital role in everyday situations, from walking on a floor to vehicles driving on a road.
Coefficient of Friction
The coefficient of friction is a dimensionless number that indicates how much frictional force exists between two surfaces. It is a key factor in determining how easily an object can slide across a surface. The coefficient provides a comparative measure of friction, with higher values indicating more friction.

In this scenario, we use the coefficient of kinetic friction, which applies when objects are moving relative to each other. To find it, we use the formula \( \mu = \frac{F_f}{F_n} \), where \(F_f\) is the friction force and \(F_n\) is the normal force. With the exercise's values, it results in a coefficient of kinetic friction of 0.25.

Understanding this coefficient allows us to predict how surfaces will interact under varying conditions. With the safe, a coefficient of 0.25 signifies a moderate level of friction, needing a notable but not excessive amount of effort to keep the safe sliding steadily. This concept is widely applicable, from machinery to transportation, and emphasizes the importance of accounting for different surface interactions.

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Most popular questions from this chapter

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you think the coefficient of friction between the sled and the snow is. You've been walking at a steady \(1.5 \mathrm{m} / \mathrm{s}\), and the rope pulls up on the sled at a \(30^{\circ}\) angle. You estimate that the mass of the sled, with your friend on it, is \(60 \mathrm{kg}\) and that you're pulling with a force of \(75 \mathrm{N}\). What answer will you give?

A 500 kg piano is being lowered into position by a crane while two people steady it with ropes pulling to the sides. Bob's rope pulls to the left, \(15^{\circ}\) below horizontal, with \(500 \mathrm{N}\) of tension. Ellen's rope pulls toward the right, \(25^{\circ}\) below horizontal. a. What tension must Ellen maintain in her rope to keep the piano descending at a steady speed? b. What is the tension in the main cable supporting the piano?

A \(1.0 \mathrm{kg}\) ball hangs from the ceiling of a truck by a 1.0 -m-long string. The back of the truck, where you are riding with the ball, has no windows and has been completely soundproofed. The truck travels along an exceedingly smooth test track, and you feel no bumps or bounces as it moves. Your only instruments are a meter stick, a protractor, and a stopwatch. a. The driver tells you, over a loudspeaker, that the truck is either at rest, or it is moving forward at a steady speed of \(5 \mathrm{m} / \mathrm{s} .\) Can you determine which it is? If so, how? If not, why not? b. Next, the driver tells you that the truck is either moving forward with a steady speed of \(5 \mathrm{m} / \mathrm{s}\), or it is accelerating at \(5 \mathrm{m} / \mathrm{s}^{2} .\) Can you determine which it is? If so, how? If not, why not? c. Suppose the truck has been accelerating forward at \(5 \mathrm{m} / \mathrm{s}^{2}\) long enough for the ball to achieve a steady position. Does the ball have an acceleration? If so, what are the magnitude and direction of the ball's acceleration? d. Draw a free-body diagram that shows all forces acting on the ball as the truck accelerates. e. Suppose the ball makes a \(10^{\circ}\) angle with the vertical. If possible, determine the truck's velocity. If possible, determine the truck's acceleration.

II It's moving day, and you need to push a \(100 \mathrm{~kg}\) box up a \(20^{\circ}\) ramp into the truck. The coefficients of friction for the box on the ramp are \(\mu_{\mathrm{s}}=0.90\) and \(\mu_{\mathrm{k}}=0.60 .\) Your largest pushing force is \(1000 \mathrm{~N}\). Can you get the box into the truck without assistance if you get a running start at the ramp? If you stop on the ramp, will you be able to get the box moving again?

A football coach sits on a sled while two of his players build their strength by dragging the sled across the field with ropes. The friction force on the sled is \(1000 \mathrm{N}\) and the angle between the two ropes is \(20^{\circ} .\) How hard must each player pull to drag the coach at a steady \(2.0 \mathrm{m} / \mathrm{s} ?\)

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