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A force of 5.00 N acts through a distance of 12.0 m in the direction of the force. Find the work done.

Short Answer

Expert verified
Answer: The work done by the force over the distance is 60.0 J.

Step by step solution

01

Write down the given values and formula

We are given: - Force, F=5.00 N - Distance, d=12.0 m - Angle, θ=0 The formula for work done is: W=Fdcosθ
02

Calculate the cosine of the angle

As the angle is 0, the cosine of the angle is: cos0=1
03

Plug in the values into the formula and calculate the work done

Using the values and the cosine of the angle, the work done can be calculated as: W=(5.00 N)(12.0 m)(1) W=60.0 J So, the work done by the force over the distance is 60.0 J.

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Key Concepts

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

Understanding Force
The concept of force is central to the discussion of work done in physics. Force can be thought of as a push or pull acting upon an object due to its interaction with another object. In the context of our exercise, a force of 5.00 Newtons (N) is applied. Force is typically measured in newtons (symbol: N) in the metric system, and it can cause an object to move, stop, or change direction.
Force is a vector quantity, which means it has both a magnitude (how much) and a direction (which way). When calculating work, the direction of the force relative to the direction of movement is significant.
The formula used in this exercise to calculate work involves the component of force acting in the direction of movement, represented by the formula:
  • Work done ( W ) = Force (F) × Distance (d) × cos(θ)
Where θ is the angle of force application. In our scenario, since the force acts in the same direction as the movement, we'd focus on its effective component.
The Role of Distance
Distance is the measure of how much an object has moved, often expressed in meters (m) in the metric system. In the mentioned exercise, the object moves a distance of 12.0 meters due to the applied force.
The distance moved is a scalar quantity, meaning it only has magnitude without a direction. In the calculation of work, the entire distance is taken into account when the force acts directly along its path, or adjusted by the cosine of the angle if it's not the case.
Therefore, distance plays a crucial role in determining the work done. The longer the distance over which the force is applied, the more work is done. In this exercise, because the force direction aligns perfectly with the distance covered, the full 12.0 meters contributes to the work calculation at face value without any reduction due to angle consideration.
Angle of Force Application
The angle between the force applied and the direction of movement is fundamental in calculating the actual work done. Given in degrees, this angle defines the component of force that effectively contributes to the movement.
In our exercise, the angle ( θ ) provided is 0 degrees, meaning that the force is aligned perfectly with the direction of motion. This alignment simplifies the calculation since cos(0)=1 .
In general:
  • A 0-degree angle means the full force is used entirely for work in the direction of the motion. It results in the maximum possible work for that force and distance.
  • A 90-degree angle indicates no work is done in moving the object in that direction as the force is perpendicular to the movement.
Thus, knowing the angle of force application helps us determine how much of the applied force contributes to the effective work done.

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Most popular questions from this chapter

A flatbed truck is loaded with a stack of sacks of cement whose combined mass is 1143.5 kg. The coefficient of static friction between the bed of the truck and the bottom sack in the stack is 0.372, and the sacks are not tied down but held in place by the force of friction between the bed and the bottom sack. The truck accelerates uniformly from rest to 56.6mph in 22.9 s. The stack of sacks is 1 m from the end of the truck bed. Does the stack slide on the truck bed? The coefficient of kinetic friction between the bottom sack and the truck bed is 0.257. What is the work done on the stack by the force of friction between the stack and the bed of the truck?

How much work is done against gravity in lifting a 6.00kg weight through a distance of 20.0 cm?

The mass of a physics textbook is 3.4 kg. You pick the book up off a table and lift it 0.47 m at a constant speed of 0.27 m/s a) What is the work done by gravity on the book? b) What is the power you supplied to accomplish this task?

Two cars are moving. The first car has twice the mass of the second car but only half as much kinetic energy. When both cars increase their speed by 5.0 m/s, they then have the same kinetic energy, Calculate the original speeds of the two cars.

A small blimp is used for advertising purposes at a football game. It has a mass of 93.5 kg and is attached by a towrope to a truck on the ground. The towrope makes an angle of 53.3 downward from the horizontal, and the blimp hovers at a constant height of 19.5 m above the ground. The truck moves on a straight line for 840.5 m on the level surface of the stadium parking lot at a constant velocity of 8.90 m/s. If the drag coefficient (K in F=Kv2) is 0.500 kg/m, how much work is done by the truck in pulling the blimp (assuming there is no wind)?

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