When a person walks (assuming the feet don't slip), the ground is pushed backward by the foot while the force of static friction opposes the force and the foot remains stationary while the upper body moves. - This is what I've understood. However, the upper body moves - I believe, since the foot exerts a force backward and the person moves forwards, the work done by the body must be negative, while, since friction acts along the direction of motion, the work done by it should be positive. These two negative and positive forces must cancel each other out, so the net work done while walking is zero. Could someone please explain if this right or wrong?
123133
While walking, the work done by friction is zero. But who does the work, actually? How someone is getting displaced? This situation also arises when someone climbs without slipping or is climbing a ladder.
42918
My Mechanics textbook claims that the sum of the work by internal forces is not generally zero.  translated to English the paragraph reads:     Notice about the work by internal forces: the work by the internal forces is in general not equal to zero although the sum of the internal forces is always equal to zero. Since the work also depends on the relative displacement of the two pointmasses in the material system, the work provided by these 2 internal forces will generally not be equal to zero.   Other sources claim that the work done by internal forces is always zero.  (answer by Mark Eichenlaub)     Then I didn't do any work on the system at all since, by definition, only external forces can do work.        the net work done by internal forces is zero since they do not cause motion of the object.   To me it would seem that these sources contradict eachother. Do they? If so, which one is correct? If not, please explain why they don't contradict eachother.
64384
So the formula for work is$$ \left[\text{work}\right] ~=~ \left[\text{force}\right]  \, \times \, \left[\text{distance}\right] \,. $$  I'm trying to get an understanding of how this represents energy.  If I'm in a vacuum, and I push a block with a force of $1 \, \mathrm{N},$ it will move forwards infinitely. So as long as I wait long enough, the distance will keep increasing. This seems to imply that the longer I wait, the more work (energy) has been applied to the block.  I must be missing something, but I can't really pinpoint what it is.  It only really seems to make sense when I think of the opposite scenario: when slowing down a block that is (initially) going at a constant speed.
5000
My physics textbook says that friction mainly arises due to intermolecular attraction between atoms of the objects in contact and clashes between peaks are only significant for rough surfaces. I was wondering, if friction arises due to intermolecular attraction, why doesn't it oppose lifting a body as well(considering that the atoms are attracted to each other). Why doesn't it oppose rolling? (even when there is no slippage, if atoms of the contact patch are attracted to the atoms in the ground, there should be a force opposing movement)   Another related question: Is there an intuitive reason why(for most pairs of surfaces) kinetic friction is lower than static friction?
102620
suppose i jump from ground then work done by normal is zero as point of application  displacement is zero(work is doneby internal forces) but during the pushing part  COM has upwards displacement then shouldn't work be done if apply force on COM  ??
66525
I did a weird activity and now I am trying to figure out the physics behind it:  While sitting on a chair I kept my feet above the ground and tried to move the chair forward. I was able to. Initially, I thought that perhaps the center of mass won't be changing as I was pushed backwards. But I moved the chair for a while and saw that it had moved a considerable distance and I was still sitting on the chair. Hence the center of mass did change for sure. According to Newton's second law, there must be an external force acting on the system(me and the chair). But I am not able to figure out what force is it. I have a feeling that it is somewhere related to the fact that the system is not completely isolated but I am not sure how.     Note: I am not in contact with the ground or any other object close to me. I have lifted my feet and have my hands on my lap.
50580
Obviously, the net work done is zero, because there's no motion, but is the proper way to look at it that both the object's gravity and the table's normal force do zero work, or that one does positive work and one an equal amount of negative work?
13720
I'm unable to understand how friction causes motion in cars and also is responsible for stopping them. When it starts the impending motion of tyres is in the backward direction that's why friction is in the forward direction and the car starts moving. But when the car engine turns off, the car stops after some time, but the condition of tyres are the same as in the previous case and friction must be in the forward direction.
65676
Consider a cylinder that rolls without sliding on an inclined plane. If it's placed at the top of the plane, with its center of mass at a height $h$ from the bottom, it will have a potential energy $mgh$ (considering the bottom of the plane as the zero point). Then, the cylinder will start rolling down the plane, as its weight does work on it, causing its potential energy to be transformed into rotational and translational kinetic energy. I'm confused about the origin of the rotational kinetic energy of the cylinder. Since the static friction and the normal force are applied at the point of contact with the plane, neither of them can do work on the cylinder (which is why its mechanical energy is conserved), since the cylinder is, by constraint, not sliding. This seems to imply that the cylinder's rotational kinetic energy comes from the work done by its weight. However, I don't understand how this happens, since a rigid body's weight can be seen as being applied on its center of mass, which means it can exert no torque on it. Instead, the only torque exerted with respect to its center of mass comes from static friction. Moreover, if the inclined plane was frictionless, there would be no torque exerted on the cylinder, so it would not rotate. Yet, I don't see how this situation could be distinguished from the previous one by only looking at the cylinder's mechanical energy (since the normal force would still be doing no work). So, my question is: how is gravitational potential energy converted into rotational kinetic energy, and what is friction's role on this?
32707
I'm a class 12th student and this thought just struck me from nowhere.  Assume a situation where every human on this planet turns towards west. All of them start walking simultaneously in west direction (exactly west i.e. antiparallel to Earth's rotation and none of their paths cross each other).  As Earth spins from west to east would this activity cause the earth's rotation speed to increase?   (There is frictional force acting on the earth due to motion by nearly a billion people which should produce enough torque to increase angular velocity of the earth.)   p.s. If the magnitude of friction force not enough to rotate the earth then assume each person to weigh more than 100 kg.  Also if it does happen would earth remain in the same orbit around the sun with just increased angular velocity about axis? (or would path around the sun also change due to this? )  I would really appreciate an easy solution cause I am still in class 12th.
101989
Recently, I have been doing some research on racing and tire modelling.  While I was doing this, I encountered many curves like those shown below.  (source: ) While I understand the need of slip angles to generate lateral forces (those perpendicular to the direction of motion), the need to have slip in the direction of motion in order to produce longitudinal force has me confused. After all, wouldn't a tire with no slipping (i.e. perfect traction), be able to produce forward forces far more efficiently than one that didn't use all of that potential? So, does anyone have an explanation for this phenomenon?
119340
If I apply a straight upward(perpendicular to ground) force against gravity of $5\ \mathrm{N}$ and lift an object "A" 10 meters, then the work done is:  $$ W = F \times S = 5\ \mathrm{N} \times 10\ \mathrm{m} = 50\ \mathrm{J}$$ But if I apply the same amount of force diagonally to the ground, and again push the object "A" 10 meters in the direction of force, then again:  $$ W = F \times S = 5\ \mathrm{N} \times 10\ \mathrm{m} = 50\ \mathrm{J} $$   In cases 1 and 2, the work done is equal, but the height would be different because in case one, height is equal to displacement, but in case 2 it will be less than 10 meters obviously. Then potential energy (i.e mgh) of both objects would be different. Doesn't conflict with the equation work = energy?  Work is force times displacement ($W = F S \cos(\theta)$); where $\theta$ is angle between direction of force and displacement) and is path independent, according to most of the text books I've read so far.
25589
From Wikipedia: The work done by a constant force of magnitude F on a point that moves a displacement d in the direction of the force is the product: $$W = Fd.$$  If I lift some object from a ground, the force to be put in above equation is the gravitational force $mg$.   But while I am moving the object upwards, against the gravity, I must pull the object with greater force than $mg$, right? So the net force is $F_{myforce}-mg$ which results in object being lifted. But this is not the case. Why? Does it matter how strongly I pull the object?  What work will I do if I move this object in absence of gravity (in space)? Will work done be zero or not?
11114
Possible Duplicate:          Layman alert...I last did physics in high school, and am just trying to understand something my son is working on.   To keep a box moving across the floor at a constant velocity I need to apply a force equal and opposite to that of the  force working on it.  If I start applying a force greater than the kinetic friction force, the box will speed up. Will it keep speeding up indefinitely, or will the kinetic friction force I need to overcome increase as the box gets faster?
82212
Consider two blocks, one on top of the other on a frictionless table, with masses $m_1$ and $m_2$ respectively. There is appreciable friction between the blocks, with coefficients $\mu_s$ and $\mu_k$ for static and kinetic respectively. I'm considering the fairly routine problem of determining the maximum horizontal force $F$ (say, to the right) that can be applied to the top block so that the two blocks accelerate together.  The problem is not hard to solve symbolically. If the two blocks move together, their accelerations are the same, and the top block doesn't move with respect to the bottom block, so only static friction is in play. In a standard coordinate system (with $x$ oriented to the right), the sum of horizontal forces for the top block is  $$F-F_{sf}=m_1a$$  and for the bottom block  $$F_{sf}=m_2a$$  where $F_{sf}$ is the force of static friction. Solving for $a$ in these two expressions, and then equating them, gives  $$F=\frac{(m_1+m_2)F_{sf}}{m_2}$$  The maximum such force will therefore be achieved when $F_{sf}$ is maxed out at $\mu_s m_1g$, so  $$F_{max}=\frac{m_1}{m_2}\mu_s(m_1+m_2)g$$  I understand this solution, but conceptually I don't have a response to the following nagging question: $F_{max}$ is clearly larger than the max static friction force $\mu_sm_1g$ (because $\frac{m_1+m_2}{m_2}&gt;1$), so why doesn't the application of a force of magnitude $F_{max}$ to the top block cause kinetic friction to take over? This line of reasoning would suggest that applying a force $F$ of magnitude greater than $\mu_sm_1g$ would cause the top block to start moving with respect to the bottom block (in which case the blocks no longer accelerate together, as in the above solution). I'm at a loss, conceptually, to say what's wrong here. I suspect it has something to do with being careful about reference frames, but a clear explanation would be much appreciated.
108749
When drawing force diagrams I am really struggling to understand why we put the weight of an object as a force acting on it, rather than as a force acting on whatever else there is around it.  For example, if we have a book on a table, and are drawing two separate force diagrams for the forces acting on each object, then surely the weight of the book is a force acting on the table, not the book?  Or if you model a diagram of a person standing, it feels wrong to label the weight as acting on the person, when surely it is acting on the floor? And the force acting on the person is the reaction of the floor to the weight?  It all just feels a bit off! Please can someone offer some help to correct my intuition?  (I should add that I kind of understand that the weight of an object does act on it, insofar as gravity does pull things down...)
94533
If the displacement is in the direction of the force, then work is positive. If the displacement is in the opposite direction of the force, then work is negative. If the displacement is perpendicular to the direction of the force then work is zero.  So, the work done depends on direction. So work should be a vector quantity. Then how is work a scalar quantity?