Solving Energy Work Problems: Examples #1 & #2

In summary, when an object is moving at constant velocity, there is no work being done since the net force is zero and the direction of the force is perpendicular to the direction of the displacement. In example #1, Brenda does no work in carrying her suitcase because her force is perpendicular to the displacement. In example #2, the work done by the 55N force is equal to 55cos 37 degrees times 15 meters since the force is partly along the displacement.
  • #1
JudyyNunez
3
0
Hi, I was wondering if anyone could please help me understand when is there no work in Joules being produced when your object is moving at a constant velocity.

Example #1: Brenda carries a 5.0 kg suitcase as she walks m along a horizontal walkway to her room at a constant speed of 6.5 m/s. How much work does Brenda do in carrying her case?

Example #2: A force of magnitude 55N directed at an angle of 37 degrees above the horizontal moves a 30kg crate along a horizontal surface at constant velocity. How much work is done by this force in moving the crate a distance of 15m?

W=F*d F=ma

For example #1 Brenda should be doing zero work due to the fact that there is no acceleration involved. Therefore, the force would be zero since acceleration is needed to create force (F=ma).

Example #2, would the work produced equal 55cos 37 degrees times 15 meters? or would it be zero work since no acceleration?
 
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  • #2
#1: the Force Brenda applies is upward, but the motion (displacement) is sideways. The Force is neither along the displacement, nor opposite the displacement, so it does neither positive nor negative Work.

Forces CAUSE acceleration , not the other way around!

#2: the 55N Force is partly along the displacement ... 4/5 of the Force is along the motion.
So THAT portion of the Force does positive Work ... to have no change in the crate's KE, there must be some other Force doing the same amount of NEGATIVE Work to the crate - I wonder what that Force would be ...
 
  • #3
Hello JudyyNunez,

Welcome to Physics Forums!

JudyyNunez said:
Hi, I was wondering if anyone could please help me understand when is there no work in Joules being produced when your object is moving at a constant velocity.

Example #1: Brenda carries a 5.0 kg suitcase as she walks m along a horizontal walkway to her room at a constant speed of 6.5 m/s. How much work does Brenda do in carrying her case?

Example #2: A force of magnitude 55N directed at an angle of 37 degrees above the horizontal moves a 30kg crate along a horizontal surface at constant velocity. How much work is done by this force in moving the crate a distance of 15m?

W=F*d F=ma

For example #1 Brenda should be doing zero work due to the fact that there is no acceleration involved. Therefore, the force would be zero since acceleration is needed to create force (F=ma).
It's true that the net force is zero. Brenda's vertical force on the suitcase is equal in magnitude and opposite in direction to the force of gravity, so the net force is zero. But that's not why the work done is zero.

The reason that the work done is zero in this case has to do with the direction of the force relative to the direction of the displacement. In what direction is the force Brenda applies on the suitcase? Which direction does the suitcase travel?

What is the dot product between perpendicular vectors?

By the way, when dealing with the concept of work it is often okay and even necessary to consider each force individually. It's okay to distinguish the work done by Brenda, or the work done by gravity, or the work done by friction. You don't need to always jump to the net force. That doesn't matter so much for this example, but it will for other examples.

Example #2, would the work produced equal 55cos 37 degrees times 15 meters?
Yes, that is correct. But make sure you understand why. :wink:

or would it be zero work since no acceleration?

No, it's not zero. It's okay here to consider the individual forces involved. The question asks for the work done by the 55N force. You can ignore, for this example, the work done by the other forces, and you can also ignore the net force (for this example).
 
  • #4
I understand now, thank you to both of you for replying!
 
  • #5


In both examples, there is no work being produced because the objects are moving at a constant velocity. Work is defined as the product of force and displacement, and in both cases, the force and displacement are perpendicular to each other, resulting in zero work being done. This is because work is only done when there is a component of the force in the direction of the displacement.

In example #1, Brenda is carrying her suitcase at a constant speed, which means there is no net force acting on the suitcase. Therefore, no work is being done.

In example #2, the force is directed at an angle of 37 degrees above the horizontal, while the displacement is purely horizontal. This means that only a component of the force is in the direction of the displacement, resulting in zero work being done.

In summary, when an object is moving at a constant velocity, there is no change in its kinetic energy and therefore no work is being done. Work is only done when there is a change in the object's kinetic energy, which requires acceleration. I hope this helps clarify your understanding.
 

What is energy work and why is it important?

Energy work refers to the process of calculating the amount of work done on an object by a force. It is important because it helps us understand the relationship between force, distance, and work, which is crucial in many scientific fields such as physics and engineering.

What are the steps involved in solving energy work problems?

The steps involved in solving energy work problems are:

  1. Identify the given variables and their values.
  2. Determine the force acting on the object and its direction.
  3. Calculate the distance over which the force is applied.
  4. Use the formula W = Fd to calculate the work done.
  5. Check your answer and make sure it has the correct units.

Can you provide an example of solving an energy work problem?

Example 1: A car with a mass of 1000 kg is pushed with a force of 500 N for a distance of 10 meters. What is the work done on the car?
Given: m = 1000 kg, F = 500 N, d = 10 m
W = Fd = (500 N)(10 m) = 5000 J
Therefore, the work done on the car is 5000 Joules.

What is the difference between positive and negative work?

Positive work occurs when the force and displacement are in the same direction, while negative work occurs when they are in opposite directions. Positive work results in an increase in energy, while negative work results in a decrease in energy.

How can energy work problems be applied in real life?

Energy work problems can be applied in many real-life situations, such as calculating the amount of work done by a person lifting weights at the gym, determining the amount of work needed to move an object from one place to another, and calculating the work done by a machine in a factory. Understanding energy work is also important in the design and construction of structures, machines, and vehicles.

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