What is the total work of a pair of sliding friction forces
Publish: 2021-05-15 19:27:59
1. When on flat ground= μ Mg, on the slope f = μ Mgcosa (a is the inclined angle of the object on the inclined plane)
if the pressure is equal to gravity plus other forces, then f = μ Mg + F), and F is the other force. In a word, the sliding friction is equal to the coefficient of friction multiplied by the supporting force (that is, the pressure on other objects is only small in calculation), and the pressure on other objects can be the resultant force of multiple forces
if the pressure is equal to gravity plus other forces, then f = μ Mg + F), and F is the other force. In a word, the sliding friction is equal to the coefficient of friction multiplied by the supporting force (that is, the pressure on other objects is only small in calculation), and the pressure on other objects can be the resultant force of multiple forces
2. 1、 Sliding friction
when two objects are in direct contact, elastic force appears on the contact surface, and the contact surface is not smooth. When there is relative motion on the contact surface, the two objects exert sliding friction to each other to hinder the relative motion
1. Does sliding friction do work
from the above description of sliding friction, it is easy to draw a conclusion: sliding friction must do work. In fact, this conclusion is wrong. Although there must be relative motion between two objects with sliding friction, s in the formula for calculating work is the displacement of the stressed object relative to the ground, and there is relative motion between the two objects, but not all the two objects have displacement relative to the ground
as shown in Figure 1, objects a and B are stacked on the horizontal ground, and object a is tied to the vertical wall with a string. The contact between the two objects is not smooth, so object B is pulled out with horizontal tension. In the process of pulling out object B, the sliding friction between B and a is horizontal to the right, while the displacement of object a relative to the ground is zero, so the sliding friction between B and a does no work for a
to judge whether sliding friction does work, we must first find out which force does work on that object. The key is to see whether the displacement of the object relative to the ground in the direction of friction is zero
2. Must sliding friction do negative work
since the direction of friction is always opposite to the direction of relative motion, for example, the direction of sliding friction between a and B is always opposite to the direction of motion between B and a, it is easy to draw the wrong conclusion that the sliding friction must do negative work
to judge whether the sliding friction is doing negative work or positive work, we must first make clear which force is doing work on which object. The key is to judge whether the angle between the direction of sliding friction and its displacement direction relative to the ground is greater than, equal to or less than 90o, which corresponds to doing negative work, not doing work and doing positive work respectively
as shown in Figure 2, a long board B with unsmooth surface is placed on the smooth horizontal ground, and a small object a can be regarded as a particle slides from the left end of the board to the right end at the horizontal initial velocity vo. As shown in Fig. 3 and Fig. 4, before a leaves B, the sliding friction Fab of object a is horizontally to the left, and the displacement SA direction of a relative to the ground is to the right, so the sliding friction Fab does negative work on a; The direction of the sliding friction FBA on B is right, and the direction of the displacement sb on the ground is right. The sliding friction FBA does positive work on B
3. Is the algebraic sum of the work of a pair of sliding friction forces not zero
the forces between objects always interact with each other, and the sliding friction between two objects is no exception. For example, in Figure 2, when a applies sliding friction FBA to B, it is also affected by the sliding friction FBA of B to a, which is the reaction force of this force. According to Newton's third law, these two forces are equal in size. If their size is f, then in the above process, The work of these two forces is: respectively,. Because | SA | & gt| So wa + WB ≠ 0
there is sliding friction between two objects, and the two objects must slide relative to each other. In a period of time, their respective displacements relative to the ground must not be equal, and the algebraic sum of the work of a pair of friction forces as force and reaction force must not be equal to zero
4. Is the algebraic sum of a pair of sliding friction work equal to the increment of their mechanical energy
in the "slider model" as shown in Figure 2, let the masses of a and B be m and m respectively, and their velocities when a leaves B be VA and VB respectively. The theorem of kinetic energy is used for a and B respectively. Thus, it can be seen that the algebraic sum of this pair of friction work is equal to the increment of mechanical energy of a and B systems. At the same time, it is not difficult to see. It can also be seen that the mechanical energy lost by the system. Therefore, the loss of mechanical energy or the increase of internal energy can also be calculated by the model
because the sliding friction does negative work on a, that is to say, a does work against the sliding friction, the amount of work is equal to the reced mechanical energy of A. It can be seen from the above discussion that part of the reced mechanical energy of a is transferred to the kinetic energy of B, and part of it is converted to the internal energy of the system
therefore, in the "slider model", the algebraic sum of the work of a pair of sliding friction forces acting as force and reaction force must be equal to the increment of their mechanical energy
5. Is the amount of work done by sliding friction only related to displacement
in the problem of work done by sliding friction discussed above, the size and direction of the sliding friction involved are constant, that is, in the process of discussion, the friction is a constant force. When calculating the work of friction, the displacement of the object in the process of study is taken to calculate the work. But if the sliding friction is variable in the process of research, the situation is different. For example:
pull the slider with horizontal tension to move along the horizontal circular track with radius r for one circle, as shown in Figure 5, the known mass of the slider is m, and the dynamic friction coefficient between the block and the track is μ Find the work done by friction in this process
in the process of an object moving along a circle, the direction of friction is always opposite to the direction of motion, and the circle is tangent, which belongs to variable force, but the size remains unchanged. As shown in Figure 6, if the circular orbit is divided into infinitely many infinitesimal segments, and the friction force on each segment can be regarded as a constant force, then the work done by the friction force on each segment is respectively,,,..., and the work done by the friction force in a week
therefore, if the friction involved in the problem is not constant, when calculating the work of sliding friction, the actual path of the object motion should be considered, not just the displacement. We can not only use the formula to calculate, but also use the "micro element division" or functional relationship, energy conservation and other physical relationships used above
2. Static friction
when there is static friction between two objects, the two objects are relatively static, but they may move relative to other reference frames, and the speed is the same
6. Does static friction do work
to answer this question, we have to start from the calculation formula of work. F in the formula is the force to calculate the work. S is the displacement of the stressed object. Its size and direction are related to the selection of reference system. When calculating the work of a certain force, the reference system is the object that is stationary on the ground or relative to the ground. In specific problems, if the forced object can be simplified as a particle, then the displacement is the displacement of the particle; if it cannot be simplified as a particle, then the displacement is the displacement of the point of action of the force. In the formula is the angle between the direction of displacement and the direction of force. Force F, displacement s, cos in the formula α， As long as one of the three quantities is zero, the work will be zero, that is, no work will be done. When we talk about whether static friction does work, the premise is that static friction must exist, and then we look at s and COS α If s is zero, that is to say, the object under the action of static friction is stationary relative to the ground, the work of static friction is zero, that is, static friction does not do work. If s is not zero, but α= When the static friction is perpendicular to the moving direction of the object, the static friction does not do work. In addition, the work of static friction will not be zero, that is to say, static friction will do work
for example, we use our fingers to hold up the pen and balance the force of gravity to make the pen stationary relative to our fingers, which is the static friction force between our fingers and the pen. Its direction is vertical and upward. When the pen moves horizontally in the air, the static friction force between our fingers and the pen does not work; When the pen moves in a non horizontal direction, the static friction between the finger and the pen does work on the pen
under the premise of static friction, whether it does work depends on s or cos α Is zero
7. Does static friction do positive work or negative work
after discussing whether static friction does work, it is relatively simple to discuss this problem. It can be seen from the formula that when the angle between the direction of static friction and the direction of displacement is less than 90o, that is α& lt; At 90o, the static friction does positive work; When the angle between the direction of static friction and the direction of displacement is greater than 90o, i.e α& gt; At 90o, the static friction does negative work. As mentioned above, the work of static friction is positive if the pen is moved upward with the finger in the air; If it moves downward, the work of static friction is negative
whether static friction does positive work or negative work depends on the angle between the direction of static friction and the direction of displacement
8. Is the algebraic sum of a pair of static friction work zero
the interaction of forces between objects is mutual, that is, force and reaction, which always act on two objects equivalently, reversely and collinear, and static friction is no exception. So, is the algebraic sum of the work done by a pair of static friction forces acting on each other and reacting on each other zero? Since the static friction force is relatively static, the displacement relative to the ground is the same. The answer is yes, that is, the algebraic sum of the work of a pair of static friction forces as the acting force and reaction force must be zero
for example, if an object is placed on a horizontal ground and pulled by a horizontal pull F, the object is still stationary. At this time, the static friction between the ground and the object and the ground is F. because the displacement of the object (particle) relative to the ground is zero, the displacement of the point of action of the static friction between the object and the ground relative to the ground is also zero, of course, The work of these two forces on each other is naturally zero, and the algebraic sum of the work of the static friction of these two forces is naturally zero. For another example, place two objects a and B on the horizontal ground, and pull object a with horizontal tension F, so that the two objects a and B remain relatively static and move in a straight line together, as shown in Figure 7. In the process of motion, the displacement of two objects is the same, the static friction of a to B is equal to that of B to a, and the current direction is opposite
9
at first glance, this problem seems to be doing positive work. In fact, the static friction of the ground against people does not do work. In the process of walking, only when people's feet touch the ground can they be affected by the static friction of the ground to the feet. When the feet are lifted and moved forward, they are not affected by the static friction of the ground. From the foot touching to the ground, the action point of static friction does not move relative to the ground, that is, the displacement of the foot relative to the ground is zero, so the static friction of the ground does no work to the human. In the same way, the static friction between human and the ground does not work on the ground
since people don't do work, how can they get kinetic energy when they walk? When walking, people's feet are inclined backward to the ground. When landing, the lower limbs of the human body exert an oblique forward and upward force on the trunk, and the trunk moves forward. Therefore, this force does positive work on the human body and enables people to obtain kinetic energy. When people land and lift their feet forward, they need the contraction and relaxation of lower limb muscles, which naturally consumes the biochemical energy of the human body. Therefore, the kinetic energy that people get ring walking actually comes from the chemical energy of human body. The existence of the static friction between the ground and the human body ensures the implementation of the force of the lower limbs on the trunk when walking, and provides conditions for human muscles to exert force and work
10. Does the friction between the ground and the driving wheel do work on the car in motion
in the movement of the car, the engine is driven by the transmission
when two objects are in direct contact, elastic force appears on the contact surface, and the contact surface is not smooth. When there is relative motion on the contact surface, the two objects exert sliding friction to each other to hinder the relative motion
1. Does sliding friction do work
from the above description of sliding friction, it is easy to draw a conclusion: sliding friction must do work. In fact, this conclusion is wrong. Although there must be relative motion between two objects with sliding friction, s in the formula for calculating work is the displacement of the stressed object relative to the ground, and there is relative motion between the two objects, but not all the two objects have displacement relative to the ground
as shown in Figure 1, objects a and B are stacked on the horizontal ground, and object a is tied to the vertical wall with a string. The contact between the two objects is not smooth, so object B is pulled out with horizontal tension. In the process of pulling out object B, the sliding friction between B and a is horizontal to the right, while the displacement of object a relative to the ground is zero, so the sliding friction between B and a does no work for a
to judge whether sliding friction does work, we must first find out which force does work on that object. The key is to see whether the displacement of the object relative to the ground in the direction of friction is zero
2. Must sliding friction do negative work
since the direction of friction is always opposite to the direction of relative motion, for example, the direction of sliding friction between a and B is always opposite to the direction of motion between B and a, it is easy to draw the wrong conclusion that the sliding friction must do negative work
to judge whether the sliding friction is doing negative work or positive work, we must first make clear which force is doing work on which object. The key is to judge whether the angle between the direction of sliding friction and its displacement direction relative to the ground is greater than, equal to or less than 90o, which corresponds to doing negative work, not doing work and doing positive work respectively
as shown in Figure 2, a long board B with unsmooth surface is placed on the smooth horizontal ground, and a small object a can be regarded as a particle slides from the left end of the board to the right end at the horizontal initial velocity vo. As shown in Fig. 3 and Fig. 4, before a leaves B, the sliding friction Fab of object a is horizontally to the left, and the displacement SA direction of a relative to the ground is to the right, so the sliding friction Fab does negative work on a; The direction of the sliding friction FBA on B is right, and the direction of the displacement sb on the ground is right. The sliding friction FBA does positive work on B
3. Is the algebraic sum of the work of a pair of sliding friction forces not zero
the forces between objects always interact with each other, and the sliding friction between two objects is no exception. For example, in Figure 2, when a applies sliding friction FBA to B, it is also affected by the sliding friction FBA of B to a, which is the reaction force of this force. According to Newton's third law, these two forces are equal in size. If their size is f, then in the above process, The work of these two forces is: respectively,. Because | SA | & gt| So wa + WB ≠ 0
there is sliding friction between two objects, and the two objects must slide relative to each other. In a period of time, their respective displacements relative to the ground must not be equal, and the algebraic sum of the work of a pair of friction forces as force and reaction force must not be equal to zero
4. Is the algebraic sum of a pair of sliding friction work equal to the increment of their mechanical energy
in the "slider model" as shown in Figure 2, let the masses of a and B be m and m respectively, and their velocities when a leaves B be VA and VB respectively. The theorem of kinetic energy is used for a and B respectively. Thus, it can be seen that the algebraic sum of this pair of friction work is equal to the increment of mechanical energy of a and B systems. At the same time, it is not difficult to see. It can also be seen that the mechanical energy lost by the system. Therefore, the loss of mechanical energy or the increase of internal energy can also be calculated by the model
because the sliding friction does negative work on a, that is to say, a does work against the sliding friction, the amount of work is equal to the reced mechanical energy of A. It can be seen from the above discussion that part of the reced mechanical energy of a is transferred to the kinetic energy of B, and part of it is converted to the internal energy of the system
therefore, in the "slider model", the algebraic sum of the work of a pair of sliding friction forces acting as force and reaction force must be equal to the increment of their mechanical energy
5. Is the amount of work done by sliding friction only related to displacement
in the problem of work done by sliding friction discussed above, the size and direction of the sliding friction involved are constant, that is, in the process of discussion, the friction is a constant force. When calculating the work of friction, the displacement of the object in the process of study is taken to calculate the work. But if the sliding friction is variable in the process of research, the situation is different. For example:
pull the slider with horizontal tension to move along the horizontal circular track with radius r for one circle, as shown in Figure 5, the known mass of the slider is m, and the dynamic friction coefficient between the block and the track is μ Find the work done by friction in this process
in the process of an object moving along a circle, the direction of friction is always opposite to the direction of motion, and the circle is tangent, which belongs to variable force, but the size remains unchanged. As shown in Figure 6, if the circular orbit is divided into infinitely many infinitesimal segments, and the friction force on each segment can be regarded as a constant force, then the work done by the friction force on each segment is respectively,,,..., and the work done by the friction force in a week
therefore, if the friction involved in the problem is not constant, when calculating the work of sliding friction, the actual path of the object motion should be considered, not just the displacement. We can not only use the formula to calculate, but also use the "micro element division" or functional relationship, energy conservation and other physical relationships used above
2. Static friction
when there is static friction between two objects, the two objects are relatively static, but they may move relative to other reference frames, and the speed is the same
6. Does static friction do work
to answer this question, we have to start from the calculation formula of work. F in the formula is the force to calculate the work. S is the displacement of the stressed object. Its size and direction are related to the selection of reference system. When calculating the work of a certain force, the reference system is the object that is stationary on the ground or relative to the ground. In specific problems, if the forced object can be simplified as a particle, then the displacement is the displacement of the particle; if it cannot be simplified as a particle, then the displacement is the displacement of the point of action of the force. In the formula is the angle between the direction of displacement and the direction of force. Force F, displacement s, cos in the formula α， As long as one of the three quantities is zero, the work will be zero, that is, no work will be done. When we talk about whether static friction does work, the premise is that static friction must exist, and then we look at s and COS α If s is zero, that is to say, the object under the action of static friction is stationary relative to the ground, the work of static friction is zero, that is, static friction does not do work. If s is not zero, but α= When the static friction is perpendicular to the moving direction of the object, the static friction does not do work. In addition, the work of static friction will not be zero, that is to say, static friction will do work
for example, we use our fingers to hold up the pen and balance the force of gravity to make the pen stationary relative to our fingers, which is the static friction force between our fingers and the pen. Its direction is vertical and upward. When the pen moves horizontally in the air, the static friction force between our fingers and the pen does not work; When the pen moves in a non horizontal direction, the static friction between the finger and the pen does work on the pen
under the premise of static friction, whether it does work depends on s or cos α Is zero
7. Does static friction do positive work or negative work
after discussing whether static friction does work, it is relatively simple to discuss this problem. It can be seen from the formula that when the angle between the direction of static friction and the direction of displacement is less than 90o, that is α& lt; At 90o, the static friction does positive work; When the angle between the direction of static friction and the direction of displacement is greater than 90o, i.e α& gt; At 90o, the static friction does negative work. As mentioned above, the work of static friction is positive if the pen is moved upward with the finger in the air; If it moves downward, the work of static friction is negative
whether static friction does positive work or negative work depends on the angle between the direction of static friction and the direction of displacement
8. Is the algebraic sum of a pair of static friction work zero
the interaction of forces between objects is mutual, that is, force and reaction, which always act on two objects equivalently, reversely and collinear, and static friction is no exception. So, is the algebraic sum of the work done by a pair of static friction forces acting on each other and reacting on each other zero? Since the static friction force is relatively static, the displacement relative to the ground is the same. The answer is yes, that is, the algebraic sum of the work of a pair of static friction forces as the acting force and reaction force must be zero
for example, if an object is placed on a horizontal ground and pulled by a horizontal pull F, the object is still stationary. At this time, the static friction between the ground and the object and the ground is F. because the displacement of the object (particle) relative to the ground is zero, the displacement of the point of action of the static friction between the object and the ground relative to the ground is also zero, of course, The work of these two forces on each other is naturally zero, and the algebraic sum of the work of the static friction of these two forces is naturally zero. For another example, place two objects a and B on the horizontal ground, and pull object a with horizontal tension F, so that the two objects a and B remain relatively static and move in a straight line together, as shown in Figure 7. In the process of motion, the displacement of two objects is the same, the static friction of a to B is equal to that of B to a, and the current direction is opposite
9
at first glance, this problem seems to be doing positive work. In fact, the static friction of the ground against people does not do work. In the process of walking, only when people's feet touch the ground can they be affected by the static friction of the ground to the feet. When the feet are lifted and moved forward, they are not affected by the static friction of the ground. From the foot touching to the ground, the action point of static friction does not move relative to the ground, that is, the displacement of the foot relative to the ground is zero, so the static friction of the ground does no work to the human. In the same way, the static friction between human and the ground does not work on the ground
since people don't do work, how can they get kinetic energy when they walk? When walking, people's feet are inclined backward to the ground. When landing, the lower limbs of the human body exert an oblique forward and upward force on the trunk, and the trunk moves forward. Therefore, this force does positive work on the human body and enables people to obtain kinetic energy. When people land and lift their feet forward, they need the contraction and relaxation of lower limb muscles, which naturally consumes the biochemical energy of the human body. Therefore, the kinetic energy that people get ring walking actually comes from the chemical energy of human body. The existence of the static friction between the ground and the human body ensures the implementation of the force of the lower limbs on the trunk when walking, and provides conditions for human muscles to exert force and work
10. Does the friction between the ground and the driving wheel do work on the car in motion
in the movement of the car, the engine is driven by the transmission
3. Sliding friction means that there is relative motion between two objects. So if you want to calculate the energy consumed by friction, you use relative displacement.
4. The calculation method of friction force is different between sliding friction force and static friction force. When calculating the friction force, we must distinguish between static friction force and sliding friction force= μ FN calculation. (1) according to the force on the object, calculate the positive pressure on the object FN. (2) according to the formula F= μ 2. Two force balance method: the object is in a state of equilibrium (uniform velocity, static positive), and it is solved according to the two force balance condition. 2. Calculation of static friction. 1. Balance condition method is used to solve the static friction. As shown in figure 3-3-19, a static object is placed on the horizontal plane. When a person pushes the object with horizontal thrust, the object is still, According to the balance condition of the two forces, the static friction is equal to the thrust. 2. The static friction is a passive force, which changes with the change of the external force; F ≤ Fmax is independent of the positive pressure between objects, so the formula F cannot be used= μ 3. Maximum static friction Fmax The value is equal to the external force along the contact surface when the object is just sliding, which is the maximum value of static friction and greater than sliding friction. Generally speaking, they are equal, and the maximum static friction is Fmax.. The formula is Fmax= μ Static FN, μ Static is the static friction coefficient and specific dynamic friction coefficient between objects μ As shown in figure 3-3-20, the dynamic friction coefficient between the object and the ground is 0.5, and the maximum static friction and sliding friction are regarded as equal, giving the object a horizontal thrust. (G is taken as 10N / kg) (L) when the thrust is 5N, how much is the friction between the ground and the object 2) When the thrust is 12n, how much is the friction force between the ground and the object? Comments: when calculating the friction force, we should first
5. The total work done by a pair of interacting sliding friction forces on an object system depends on the path
6. For example:
when an object slides to a stop on a rough surface,
the friction between the object and the ground is the force and reaction,
but the object does negative work under the friction, but the ground does not move, and does no work on the ground friction.
when an object slides to a stop on a rough surface,
the friction between the object and the ground is the force and reaction,
but the object does negative work under the friction, but the ground does not move, and does no work on the ground friction.
7. The sum of work done by a pair of sliding friction forces is not zero. From the energy point of view, if there is friction between two objects, there must be heat, then the mechanical energy of the two objects must be reced, that is to say, the sum of work done by a pair of sliding friction is negative
8. Part of the work of friction heat generation is consumed in heating
9. Thrust and friction are not a pair of sliding friction
10. The work done by sliding friction usually proces heat, so the total energy is transformed from mechanical energy to heat energy in the process of doing work, and the increment of useful work is negative
if you see such a statement, what people want to say is that there is sliding friction in a system, then if the sliding friction does work, the sum of the useful work done by a pair of sliding friction is negative
because sliding friction must generate heat, when calculating the mechanical energy conservation of the whole system, a pair of sliding friction can not be calculated as conservative internal force, but the corresponding idle work done by sliding friction should be dected
for example, there is a large enough static board on the smooth ground, a small iron block on the board has an initial velocity, and there is relative sliding friction between the small iron block and the board. At the beginning of the interaction between the small iron block and the large board, the mechanical energy of the whole system is only the kinetic energy of the small iron block. At the end of the equilibrium, the small iron block stays on the board, and momentum is conserved, But at this time, the sum of the kinetic energy of the small iron block to increase the board is smaller than that of the previous small iron block. In this process, only a pair of sliding friction forces do work, and the work done by the sliding friction forces proces idle work. Therefore, it can be said that the algebraic sum of the work done by a pair of sliding friction forces is negative. Of course, if the part of idle work that generates heat is included, the total work is 0, It's just that we didn't calculate the work that generated the heat
this example can directly replace the data. Suppose that the mass of the board and the small iron block is m, and the initial velocity of the small iron block is V, then
the initial kinetic energy
E1 = 1 / 2 * m * V ^ 2
in equilibrium, the velocity becomes half, and the kinetic energy of the system is
E2 = 1 / 2 * 2m * (V / 2) ^ 2 = 1 / 4 * m * V ^ 2
obviously, the kinetic energy of the system becomes half of the original, and in this process, half of the kinetic energy is converted into heat energy.
if you see such a statement, what people want to say is that there is sliding friction in a system, then if the sliding friction does work, the sum of the useful work done by a pair of sliding friction is negative
because sliding friction must generate heat, when calculating the mechanical energy conservation of the whole system, a pair of sliding friction can not be calculated as conservative internal force, but the corresponding idle work done by sliding friction should be dected
for example, there is a large enough static board on the smooth ground, a small iron block on the board has an initial velocity, and there is relative sliding friction between the small iron block and the board. At the beginning of the interaction between the small iron block and the large board, the mechanical energy of the whole system is only the kinetic energy of the small iron block. At the end of the equilibrium, the small iron block stays on the board, and momentum is conserved, But at this time, the sum of the kinetic energy of the small iron block to increase the board is smaller than that of the previous small iron block. In this process, only a pair of sliding friction forces do work, and the work done by the sliding friction forces proces idle work. Therefore, it can be said that the algebraic sum of the work done by a pair of sliding friction forces is negative. Of course, if the part of idle work that generates heat is included, the total work is 0, It's just that we didn't calculate the work that generated the heat
this example can directly replace the data. Suppose that the mass of the board and the small iron block is m, and the initial velocity of the small iron block is V, then
the initial kinetic energy
E1 = 1 / 2 * m * V ^ 2
in equilibrium, the velocity becomes half, and the kinetic energy of the system is
E2 = 1 / 2 * 2m * (V / 2) ^ 2 = 1 / 4 * m * V ^ 2
obviously, the kinetic energy of the system becomes half of the original, and in this process, half of the kinetic energy is converted into heat energy.
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