Work and Energy on a Slope - How Does a Block Move Up with Zero Net Work?

Click For Summary
A block on a slope is pulled up at constant speed, leading to confusion about net work and energy changes. The net work done on the block is zero because its kinetic energy remains constant, but it still gains potential energy as it moves up. This is explained by recognizing that the work done by the tension in the string equals the negative work done by gravity. The relationship between initial energy, work done, and final energy must account for gravitational potential energy without double counting. Ultimately, while the block's kinetic energy does not change, the work done by the pulling force translates into an increase in potential energy.
JackyCheukKi
Messages
1
Reaction score
1
Homework Statement
Work energy theorem please help me
Relevant Equations
energy initial(of one object) + net work done on the object = energy final??
Guys, I have a problem that really needs you guys to help, I know it is a stupid question but please bear with me:

Context:
You have a block on a slope(has friction) you use a string to pull the block up with constant speed.

Problem:
So according to the network theorem, the work net is equal to the change in kinetic energy, and here we can see that the kinetic energy remains the same and the net work should be zero. But my problem is if the net work is zero, how the heck did the block move up the slope?? if it moves up the slope, it gains POTENTIAL ENERGY right??

isn't Energy(initial) + net work = Energy(final) ?
 
Physics news on Phys.org
One must be careful not to count the effects of gravity twice. Either you include the work done by gravity as a force or the change in gravitational potential energy, but not both. (Gravitational PE already includes work done by gravity.)

There are several ways to look at the Work-KE theorem. If you include the work done by ALL forces, including gravity, then no net work is done on the block as it is pulled up the incline. Thus the KE doesn't change. So what? While there's no net work done, whoever is pulling the string is certainly doing work!

On the other hand, you can also view it in terms of energy: Initial energy + work done = final energy. If your energy term includes gravitational PE, then the "work done" means the work done by all forces except gravity. That "work done" will not be zero, since PE increases.
 
  • Like
Likes Delta2, Lnewqban, Steve4Physics and 2 others
I have a slightly different view from @Doc Al's.

By definition, the net work done on a rigid body is the work done by the net force. Thus, it is equal to the gain in KE. Always. The gain in PE is not considered work done on the body; rather it is work done on the system consisting of the two gravitationally attracted bodies.

You are confusing net work done on the body with the work done by just one applied force, the tension in the string. While the speed is constant, ignoring friction, that is equal and opposite to the work done by gravity.

The precise wording in post #1 suggests the body starts at rest, so you do have to do some net work on it to get it going, though this can be arbitrarily small, depending how much of a hurry you are in. If it also finishes at rest, and there's no friction, you will stop pulling a little before that.
 
Yes the block gains potential energy but doesn't gain kinetic energy because the net work is zero. Because the net work is zero the work of the tension equals minus the work of the weight. As you probably know the gain in potential energy equals also minus the work of the weight, hence it is equal to the work of the tension. To summarize it:
  • Gain in kinetic energy=net work=0
  • Gain in potential energy=-work of weight=work of tension
 
Last edited:
Beams of electrons and protons move parallel to each other in the same direction. They ______. a. attract each other. b. repel each other. c. neither attract nor repel. d. the force of attraction or repulsion depends upon the speed of the beams. This is a previous-year-question of CBSE Board 2023. The answer key marks (b) as the right option. I want to know why we are ignoring Coulomb's force?