# How much work is done when pushing against a brick wall?

So this is one of the questions on my physics assessment and I would like to know how to calculate how much work is done when pushing against a brick wall.

My teacher has told me the calculation is: weight(kg)x10 (this gives me the force)x distance.

But what is the weight of and the distance of? As when you push against a brick wall the wall doesn't move and you can stop yourself from moving.

Please answer as soon as possible, also I'm kind of younger so could you explain clearly(sorry)

Emsee

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You have all of the components you need to solve this problem. As you say, the wall doesn't move; therefore, the distance is... And the work performed is... – AdamRedwine Feb 14 '13 at 16:25
@AdamRedwine A valid question: is this a duplicate of another question we've seen? This is one of those that every physics student will ask and every basic introduction class should teach, but if its the first time our site has seen the question, shouldn't we provide an answer and then call it done? – Alan Rominger Feb 14 '13 at 16:45
physicsforums.com and Yahoo Q&A is a good place to ask homework questions like this. – Larry Harson Feb 14 '13 at 17:33
@AlanSE, not necessarily. The forum guidelines say that questions of a homework nature should demonstrate some effort put forth toward solving the problem on the part of the poster and that the answer should not simply answer the question but should guide the poster in the right direction. Did my comment not follow those guides? – AdamRedwine Feb 14 '13 at 21:37
@AdamRedwine I'm not saying there's anything wrong with your comment, there isn't. The rules have some ambiguity in them. For instance, someone could argue that the (apparent) attempt here to apply $mgh$ was some effort put forth. I would also like to see more questioned based in deep interpretative questions as opposed to outright errors, but the problem is that the asker can't know which their question is. How to deal with these questions falls within the discretion of the community. – Alan Rominger Feb 15 '13 at 15:49

## 3 Answers

It would be twice as much work since the "machine" or object does not move or put forth more energy.

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This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. – DavePhD May 15 '14 at 1:23
Actually, I think this is technically an answer (just not a good one). It's not nonsensical or on an entirely different topic. – David Z May 15 '14 at 1:33

Emsee,

This is a bit of a trick question.

The equation you have is only correct for calculating the work done against gravity (and then only when you're near earth's surface).

The equation for calculating work is: work (joules) = Force (Newtons) * distance (meters)

Specifically, this says that the work done on an object is equal to the force applied to it multiplied by the distance that force caused it to be moved.

"My teacher has told me the calculation is: weight(kg)x10 (this gives me the force)x distance."

That equation is a specific example of a work calculation: the work done when moving an object up or down against gravity. The 10 represents acceleration due to gravity and is actually 10 meters per second per second (also known as meters per second squared). Always use units.

If the force exerted is not against gravity, it is calculated differently...but don't get started trying to figure out how to do that just yet.

Work=Force*distance

You don't know what the magnitude of your force is, but you know what direction it's in. So how much does the object that the force is applied to move in that direction?

I'll let you figure out the rest since it's a homework problem.

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For someone who has never had a physics class, I think it's fairly common to confuse the concepts of energy and force. You're speaking of a force being exerted, then asking what the energetic picture of the situation is. The answer is to simply decouple those two concepts and look at what physics has to say about them. We can formulate the problem several different ways.

• Energy is force times distance. For your problem, no matter how long you push, the wall still goes nowhere, so there is no obvious energy transfer. You're probably wasting energy that goes into body heat, but you do that anyway.

$$E = F d = F ( 0 )= 0$$

• Power is force times velocity. You also need to know that power is defined to be energy supplied per unit time. You can use this to talk about the power your car's engine is putting into the car's kinetic energy. With this perspective, we discover that no energy is dynamically being transferred at any given point in time.

$$P = F v = F (0) = 0$$

If you think about it, you are surrounded by a pressure (defined to be force per area) of $14 \text{ pounds force}/{\text{in}^2}$. To drill in the point that force on a static surface doesn't do anything to the energy balance of the world, just consider the incredible forces that the air around you is exerting on you right now. The force is balanced, evidenced by the fact that you're not currently accelerating. When you push on the brick wall, the building distributes your force through its foundation and pushes back in a similar manner.

These infinite invisible forces that permeate your world are difficult to swallow initially, simply because they are invisible.

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## protected by Qmechanic♦May 15 '14 at 22:41

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