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The brakes of a truck cause it to slow down by applying a retarding force of \(3.0 \times 10^{3} \mathrm{~N}\) to the truck over a distance of \(850 \mathrm{~m}\). What is the work done by this force on the truck? Is the work positive or negative? Why?

Short Answer

Expert verified
The work done is -2,550,000 J, which is negative as the force is retarding.

Step by step solution

01

Understanding Work Done by a Force

The work done on an object by a force is calculated using the formula: \[ W = F \times d \times \cos(\theta) \]where \(W\) is the work done, \(F\) is the magnitude of the force, \(d\) is the displacement, and \(\theta\) is the angle between the force and the direction of displacement. Since the force and displacement are in opposite directions (as the force is retarding), \(\theta = 180^\circ\).
02

Calculating the Cosine of the Angle

Since the angle \(\theta = 180^\circ\), the cosine of \(180^\circ\) is \[ \cos(180^\circ) = -1 \].This indicates that the work done by the force will have a negative value, reflecting that this is retarding or slowing down the truck.
03

Substituting Values in the Work Formula

Insert the given values for force, distance, and the angle:\[ W = 3.0 \times 10^{3} \mathrm{~N} \times 850 \mathrm{~m} \times \cos(180^\circ) \].This results in:\[ W = 3.0 \times 10^{3} \times 850 \times (-1) \].
04

Calculating the Work Done

Multiply the values to find the work done:\[ W = 3.0 \times 850 \times 10^{3} \times (-1) \]\[ W = -2,550,000 \text{ J} \]Thus, the work done by the force is \(-2,550,000 \mathrm{~J}\).
05

Interpreting the Sign of the Work Done

The negative sign of the work done indicates it is against the direction of movement, meaning it is done by a retarding force. This confirms that the force is slowing down the truck.

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Key Concepts

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

Retarding Force
A retarding force is a force that acts in the opposite direction to the motion of an object. It's what we commonly refer to as friction or drag, but in this case, it can also include forces like braking in vehicles.
This force slows down or stops the object from continuing on its path, much like the brakes on a truck. For example, on our truck problem, the retarding force provided by its brakes is crucial in bringing the truck to a stop or reducing its speed. The force given is substantial at \(3.0 \times 10^{3} \text{ N}\), meaning it's strong enough to considerably impact the truck's motion over the distance needed.
  • Retarding force opposes motion
  • Always acts against the direction of travel
  • Example: Brakes apply a retarding force
Understanding how retarding forces work can help in solving problems related to motion and energy dissipation in dynamics.
Work Formula
To determine the work done by a force, you can use the work formula:
\[ W = F \times d \times \cos(\theta) \]
This formula allows us to calculate how much energy is transferred by the force over a certain distance.
  • \(W\) represents the work done measured in Joules (J).
  • \(F\) is the force applied, in Newons (N).
  • \(d\) is the displacement, or the distance over which force is applied, in meters (m).
  • \(\theta\) is the angle between the force and the direction of movement.
Work can be positive, negative, or zero depending on the interaction between the direction of the force and displacement. The versatility of this formula makes it applicable in a variety of contexts, particularly when dealing with physical scenarios like braking.
Negative Work
Negative work occurs when the force applied to an object is opposite to its displacement. This means the force is taking energy away from the object instead of adding to it, which is typical in braking scenarios, where the object slows down.
In our truck example, the calculated work is \(-2,550,000 \text{ J}\).
The negative sign indicates that the energy from the brakes is acting against the truck's motion.
  • Negative work reduces energy from the system
  • Commonly occurs with retarding forces such as brakes
  • Reflects energy loss or conversion (e.g., into heat)
This concept of negative work helps explain how energy is managed and transformed in physical systems, particularly when slowing or stopping motion.
Cosine of Angle
The cosine of the angle is crucial in determining the directionality of the work done. In the work formula \(W = F \times d \times \cos(\theta)\), the term \(\cos(\theta)\) effectively shows how the force aligns with the direction of displacement.
When the angle \(\theta\) is \(180^\circ\), like in retarding forces, the cosine of that angle is \(-1\):
\[ \cos(180^\circ) = -1 \]
This means any work done by the force in this situation is negative, as the force direction is opposite.
  • \(\theta = 0^\circ\) means full positive work
  • \(\theta = 90^\circ\) means no work done
  • \(\theta = 180^\circ\) means maximum negative work
Understanding the cosine of the angle helps predict and verify the signs of work, crucial in solving physics problems related to forces and motion.

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Most popular questions from this chapter

One of the new events in the 2002 Winter Olympics was the sport of skeleton (see the photo). Starting at the top of a steep, icy track, a rider jumps onto a sled (known as a skeleton) and proceeds-belly down and head first-to slide down the track. The track has fifteen turns and drops \(104 \mathrm{~m}\) in elevation from top to bottom. (a) In the absence of nonconservative forces, such as friction and air resistance, what would be the speed of a rider at the bottom of the track? Assume that the speed of the rider at the beginning of the run is relatively small and can be ignored. (b) In reality, the best riders reach the bottom with a speed of \(35.8 \mathrm{~m} / \mathrm{s}\) (about \(80 \mathrm{mi} / \mathrm{h}\) ). How much work is done on an \(86.0-\mathrm{kg}\) rider and skeleton by nonconservative forces?

At a carnival, you can try to ring a bell by striking a target with a \(9.00\) -kg hammer. In response, a \(0.400-\mathrm{kg}\) metal piece is sent upward toward the bell, which is \(5.00 \mathrm{~m}\) above. Suppose that \(25.0 \%\) of the hammer's kinetic energy is used to do the work of sending the metal piece upward. How fast must the hammer be moving when it strikes the target so that the bell just barely rings?

A \(2.40 \times 10^{2}-\mathrm{N}\) force is pulling an 85.0 -kg refrigerator across a horizontal surface. The force acts at an angle of \(20.0^{\circ}\) above the surface. The coefficient of kinetic friction is 0.200 , and the refrigerator moves a distance of \(8.00 \mathrm{~m}\). Find (a) the work done by the pulling force, and (b) the work done by the kinetic frictional force.

A 47.0 -g golf ball is driven from the tee with an initial speed of \(52.0 \mathrm{~m} / \mathrm{s}\) and rises to a height of \(24.6 \mathrm{~m}\). (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b)What is its speed when it is \(8.0 \mathrm{~m}\) below its highest point?

A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates. The crest of the second hill is circular, with a radius of \(r=36 \mathrm{~m} .\) Neglect friction and air resistance. What must be the height \(h\) of the first hill so that the skier just loses contact with the snow at the crest of the second hill?

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