Solving Energy Work Problems: Examples #1 & #2

[JudyyNunez]

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?

 

[lightgrav]

#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 ...

 

[collinsmark]

Hello JudyyNunez,

Welcome to Physics Forums!

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.

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).

 

[JudyyNunez]

I understand now, thank you to both of you for replying!