Calculating Work and Energy in a Box on a Floor: A Quick Problem

In summary, The person is pushing a box on a floor with a 250 N force, which results in a constant speed of the box over 12.75 m. The quantity of work done by the person on the box is 3187.5 Joules. The quantity of work done by friction is -3187.5 Joules. The total work done is 0, as it is the sum of the work done by the person and the work done by friction. There may be a slight discrepancy in the calculations for part (c), but the overall process appears to be correct.
  • #1
ThatDude
33
0

Homework Statement



A person is pushing a box on a floor with a 250 N force. The box is moving at a constant speed over 12.75 m.

(a) What quantity of work does the person do on the box?
(b) What quantity of word does friction do?
(c) What is the total work done?

2. The attempt at a solution

(a) W = F * d * cosx = (250 N)(12.75 m)(cos0) = 3187.5 Joules

(b) W = F * d * cosx = (250 N)(12.75 m)(cos180) = -3187.5 Joules

(c) Total work = Wperson + Wfriction = 3187.5 J + (-3187.5 J) = 0
 
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  • #2
Do you have a question?
Your working seems correct.(I don't have my calculator at the moment)
 
  • #3
I was wondering if my process for (c) was correct.
 
  • #4
ThatDude said:
I was wondering if my process for (c) was correct.
Probably not correct, but now I'm unsure of whether there is a strict convention to adhere to on the topic of friction. Were I to answer this as a test, I would have given the one identical answer to all 3 parts.

I wouldn't like to say zero total work has been done, because I can see that heat has been added to the system as a result of the friction. (So I'm tipping that maybe we say the work done by friction is zero.)

We'd best wait on to see what others say.
 
Last edited:
  • #5
ThatDude said:

Homework Statement



A person is pushing a box on a floor with a 250 N force. The box is moving at a constant speed over 12.75 m.

(a) What quantity of work does the person do on the box?
(b) What quantity of word does friction do?
(c) What is the total work done?

(b) is a negative number.
(c) = (a) + (b)
 
  • #6
I can see why work done by friction is considered negative. W=F.s, so when motion is in a direction opposite to that of F then one of the terms will have a negative sign. For the block, though, where movement is in the same direction as F then their product will be positive, and positive work is done. Thanks rude man.
 
  • #7
That means all the answers in the OP are correct
 
  • #8
adjacent said:
That means all the answers in the OP are correct

EDIT: sorry, hadn't noticed the OP's answers changed. Or maybe I misread them.

In any case, yes, they are now correct if he did the math correctly.
 
Last edited:
  • #9
The confusion was mine. OP was correct from the start.
 

1. What is a quick work and energy problem?

A quick work and energy problem is a type of physics problem that involves calculating the work and energy involved in a given situation or scenario. It typically requires the use of formulas and equations to solve.

2. How do I solve a quick work and energy problem?

To solve a quick work and energy problem, you will need to identify the relevant variables and plug them into the appropriate formulas. Make sure to pay attention to units and use the correct equations for the type of problem you are solving.

3. What are the units of work and energy?

Work is typically measured in joules (J), while energy is measured in either joules (J) or electronvolts (eV). It is important to keep track of units when solving work and energy problems.

4. How is work related to energy?

Work and energy are closely related concepts in physics. Work is equal to the change in energy, and energy is the capacity of an object to do work. In other words, work is a way to transfer energy from one object to another.

5. What are some real-life examples of work and energy?

Some real-life examples of work and energy include lifting a heavy object, pushing a car, and riding a bike. In all of these situations, work is being done and energy is being transferred or transformed.

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