You are already familiar with mechanical work (work of force) from the basic school physics course. Let us recall the definition given there mechanical work for the following cases.
If the force is directed in the same direction as the movement of the body, then the work done by the force
In this case, the work done by the force is positive.
If the force is directed opposite to the movement of the body, then the work done by the force
In this case, the work done by the force is negative.
If the force f_vec is directed perpendicular to the displacement s_vec of the body, then the work done by the force is zero:
Job - scalar quantity. The unit of work is called the joule (symbol: J) in honor of the English scientist James Joule, who played important role in the discovery of the law of conservation of energy. From formula (1) it follows:
1 J = 1 N * m.
1. A block weighing 0.5 kg was moved along the table 2 m, applying an elastic force of 4 N to it (Fig. 28.1). The coefficient of friction between the block and the table is 0.2. What is the work acting on the block?
a) gravity m?
b) normal reaction forces?
c) elastic forces?
d) sliding friction forces tr?
The total work done by several forces acting on a body can be found in two ways:
1. Find the work of each force and add up these works, taking into account the signs.
2. Find the resultant of all forces applied to the body and calculate the work of the resultant.
Both methods lead to the same result. To make sure of this, go back to the previous task and answer the questions in task 2.
2. What is it equal to:
a) the sum of the work done by all forces acting on the block?
b) the resultant of all forces acting on the block?
c) work resultant? IN general case(when the force f_vec is directed under arbitrary angle to the displacement s_vec) the definition of the work of force is as follows.
Job A constant force is equal to the product of the force modulus F by the displacement modulus s and the cosine of the angle α between the direction of force and the direction of displacement:
A = Fs cos α (4)
3. Show what general definition The work follows to the conclusions shown in the following diagram. Formulate them verbally and write them down in your notebook.
4. A force is applied to a block located on the table, the modulus of which is 10 N. Why angle is equal between this force and the movement of the block, if when moving the block along the table by 60 cm, this force did the work: a) 3 J; b) –3 J; c) –3 J; d) –6 J? Make explanatory drawings.
2. Work of gravity
Let a body of mass m move vertically from the initial height h n to the final height h k.
If the body moves downwards (h n > h k, Fig. 28.2, a), the direction of movement coincides with the direction of gravity, therefore the work of gravity is positive. If the body moves upward (h n< h к, рис. 28.2, б), то работа силы тяжести отрицательна.
In both cases, the work done by gravity
A = mg(h n – h k). (5)
Let us now find the work done by gravity when moving at an angle to the vertical.
5. A small block of mass m slid along an inclined plane of length s and height h (Fig. 28.3). The inclined plane makes an angle α with the vertical.
a) What is the angle between the direction of gravity and the direction of movement of the block? Make an explanatory drawing.
b) Express the work of gravity in terms of m, g, s, α.
c) Express s in terms of h and α.
d) Express the work of gravity in terms of m, g, h.
e) What is the work done by gravity when the block moves upward along the entire same plane?
Having completed this task, you are convinced that the work of gravity is expressed by formula (5) even when the body moves at an angle to the vertical - both down and up.
But then formula (5) for the work of gravity is valid when the body moves along any trajectory, because any trajectory (Fig. 28.4, a) can be represented as a set of small “ inclined planes"(Fig. 28.4, b). 
Thus,
the work done by gravity when moving along any trajectory is expressed by the formula
A t = mg(h n – h k),
where h n is the initial height of the body, h k is its final height.
The work done by gravity does not depend on the shape of the trajectory.
For example, the work of gravity when moving a body from point A to point B (Fig. 28.5) along trajectory 1, 2 or 3 is the same. From here, in particular, it follows that the force of gravity when moving along a closed trajectory (when the body returns to the starting point) is equal to zero. 
6. A ball of mass m, hanging on a thread of length l, was deflected by 90º, keeping the thread taut, and released without a push.
a) What is the work done by gravity during the time during which the ball moves to the equilibrium position (Fig. 28.6)?
b) What is the work done by the elastic force of the thread during the same time?
c) What is the work done by the resultant forces applied to the ball during the same time?
3. Work of elastic force
When the spring returns to an undeformed state, the elastic force always does positive work: its direction coincides with the direction of movement (Fig. 28.7). 
Let's find the work done by the elastic force.
The modulus of this force is related to the modulus of deformation x by the relation (see § 15)
The work done by such a force can be found graphically.
Let us first note that the work done by a constant force is numerically equal to the area of the rectangle under the graph of force versus displacement (Fig. 28.8). 
Figure 28.9 shows a graph of F(x) for the elastic force. Let us mentally divide the entire movement of the body into such small intervals that at each of them the force can be considered constant. 
Then the work on each of these intervals is numerically equal to the area of the figure under the corresponding section of the graph. All work is equal to the sum of work in these areas.
Consequently, in this case, the work is numerically equal to the area of the figure under the graph of the dependence F(x). 
7. Using Figure 28.10, prove that
the work done by the elastic force when the spring returns to its undeformed state is expressed by the formula
A = (kx 2)/2. (7)
8. Using the graph in Figure 28.11, prove that when the spring deformation changes from x n to x k, the work of the elastic force is expressed by the formula
From formula (8) we see that the work of the elastic force depends only on the initial and final deformation of the spring. Therefore, if the body is first deformed and then returns to its initial state, then the work of the elastic force is zero. Let us recall that the work of gravity has the same property.
9. B starting moment the stretch of a spring with a stiffness of 400 N/m is 3 cm. The spring is stretched by another 2 cm.
a) What is the final deformation of the spring?
b) What is the work done by the elastic force of the spring?
10. At the initial moment, a spring with a stiffness of 200 N/m is stretched by 2 cm, and at the final moment it is compressed by 1 cm. What is the work done by the elastic force of the spring?
4. Work of friction force
Let the body slide along a fixed support. The sliding friction force acting on the body is always directed opposite to the movement and, therefore, the work of the sliding friction force is negative in any direction of movement (Fig. 28.12). 
Therefore, if you move the block to the right, and the peg the same distance to the left, then, although it will return to starting position, the total work done by the sliding friction force will not be equal to zero. This is the most important difference the work of the sliding friction force from the work of gravity and elasticity. Let us recall that the work done by these forces when moving a body along a closed trajectory is zero.
11. A block with a mass of 1 kg was moved along the table so that its trajectory turned out to be a square with a side of 50 cm.
a) Has the block returned to its starting point?
b) What is the total work done by the frictional force acting on the block? The coefficient of friction between the block and the table is 0.3.
5.Power
Often it is not only the work being done that is important, but also the speed at which the work is being done. It is characterized by power.
Power P is the ratio of the work done A to the time period t during which this work was done:
(Sometimes power in mechanics is denoted by the letter N, and in electrodynamics by the letter P. We find it more convenient to use the same designation for power.)
The unit of power is the watt (symbol: W), named after English inventor James Watt. From formula (9) it follows that
1 W = 1 J/s.
12. What power does a person develop by uniformly lifting a bucket of water weighing 10 kg to a height of 1 m for 2 s?
It is often convenient to express power not through work and time, but through force and speed.
Let's consider the case when the force is directed along the displacement. Then the work done by the force A = Fs. Substituting this expression into formula (9) for power, we obtain:
P = (Fs)/t = F(s/t) = Fv. (10)
13. A car is traveling on a horizontal road at a speed of 72 km/h. At the same time, its engine develops a power of 20 kW. What is the force of resistance to the movement of the car?
Clue. When a car is moving on a horizontal road with constant speed, the traction force is equal in magnitude to the resistance force to the movement of the car.
14. How long will it take to uniformly lift a concrete block weighing 4 tons to a height of 30 m if the power of the crane motor is 20 kW and the efficiency of the electric motor of the crane is 75%?
Clue. Electric motor efficiency equal to the ratio work on lifting loads for engine operation. 
Additional questions and tasks
15. A ball with a mass of 200 g was thrown from a balcony with a height of 10 and an angle of 45º to the horizontal. Reaching in flight maximum height 15 m, the ball fell to the ground.
a) What is the work done by gravity when lifting the ball?
b) What is the work done by gravity when the ball is lowered?
c) What is the work done by gravity during the entire flight of the ball?
d) Is there any extra data in the condition?
16. A ball with a mass of 0.5 kg is suspended from a spring with a stiffness of 250 N/m and is in equilibrium. The ball is raised so that the spring becomes undeformed and released without a push.
a) To what height was the ball raised?
b) What is the work done by gravity during the time during which the ball moves to the equilibrium position?
c) What is the work done by the elastic force during the time during which the ball moves to the equilibrium position?
d) What is the work done by the resultant of all forces applied to the ball during the time during which the ball moves to the equilibrium position?
17. A sled weighing 10 kg slides down without initial speed with snowy mountain with an inclination angle α = 30º and travel a certain distance along horizontal surface(Fig. 28.13). The coefficient of friction between the sled and snow is 0.1. The length of the base of the mountain is l = 15 m. 
a) What modulus is equal frictional forces when the sled moves on a horizontal surface?
b) What is the work done by the friction force when the sled moves along a horizontal surface over a distance of 20 m?
c) What is the magnitude of the friction force when the sled moves along the mountain?
d) What is the work done by the friction force when lowering the sled?
e) What is the work done by gravity when lowering the sled?
f) What is the work done by the resultant forces acting on the sled as it descends from the mountain?
18. A car weighing 1 ton moves at a speed of 50 km/h. The engine develops a power of 10 kW. Gasoline consumption is 8 liters per 100 km. The density of gasoline is 750 kg/m 3, and its specific heat combustion 45 MJ/kg. What is the efficiency of the engine? Is there any extra data in the condition?
Clue. The efficiency of a heat engine is equal to the ratio of the work performed by the engine to the amount of heat released during fuel combustion.
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Work, energy, power
Forces cause either the acceleration of a body (dynamic action) or a change in its shape (static action).
If a force moves a body a certain distance, then it does work on the body.
Job= Force x Displacement.
At F = const(in case of constant force during movement) A = Fs, in case variable force– integral of force over displacement A = .
Power– the ratio of the work performed to the time during which it was performed:
Power = Work / Time .
Instantaneous power is the derivative of work with respect to time: R = dA/dt. Because dA = Fds(force to move), then R = Fds/dt = Fv. Instantaneous power is equal to the product instantaneous force at instant speed.
Energy- the body's ability to do work, single measure various forms movements. Quantitative characteristics depend on the type of energy (mechanical, internal, chemical, nuclear, electromagnetic, etc.).
There are two ways of transferring motion and its corresponding energy from one body to another - in the form of work and in the form of heat (through heat exchange). For microparticles (atoms, electrons) these concepts are not applicable.
If a body moves in the direction of gravity, then work is done on the body A = Gh or A T = mg h.
To raise a body (increase the distance from the center of the Earth), work must be done on it. Work done by force F when moving against gravity (lifting the body) to a height h does not depend on the path - it depends only on how much the body can descend to a given level. This work is stored in the form of potential energy of the body (position energy) A=W n = mgh, equal work, spent on lifting the body.
This is not the total potential energy - only the increment of energy when the body rises to a height (the reference point is chosen arbitrarily). Subject to change gravitational field in height W n = m .
Potential energy called energy that depends only on relative position material points (or bodies).
Forces acting on material point(body), are called potential if the work of these forces when moving a point (body) depends only on the initial and final position of the point (body) in space and does not depend on the path of movement.
In all physical phenomena It is not the potential energy itself that is important, but its change, which determines the work done. The level of reference for changes is agreed upon in advance.
Potential energy includes position energy and elastic deformation energy.
Potential energy can be possessed not only by a system of interacting forces, but also by a single elastically deformable body (compressed spring, stretched rod). In this case, the potential energy depends on the relative position individual parts body (spring coils).
Kinetic energy body is the measure of it mechanical movement and is measured by the work that a body can do when braking to a complete stop.
From a state of rest, change in speed and path to moment t: V =at, S=Vt/2=at 2 /2.
When braking, a force acts on the body directed against its movement. Until the body comes to a complete stop under the influence of force F will do the work A: A = Fs = Fv 2 /2a = mv 2 /2.
Kinetic energy of the body K = mv 2 /2
 |
When rising to a height, potential energy accumulates W n, when falling from this height, this potential energy turned into kinetic W To. W n = W k = mgh = mv 2 /2.
Example: determination of speed using a pendulum-weight.
1. Formulation of the content model
Determine the speed of the bullet. The problem is solved using a pendulum-weight suspended on a light, rigid and freely rotating rod. Initial data - in accordance with the figure.
2. Formulation of the conceptual model
A bullet stuck in the load will tell the bullet-load system its kinetic energy, which at the moment of the greatest deviation of the rod from the vertical will completely transform into the potential energy of the system. The solution to the problem is based on the law of conservation of energy. Energy losses for heating the bullet and load, overcoming air resistance, acceleration of the rod, etc. are not taken into account.
3. Development of a mathematical model.
This transformation is described by a chain of equalities from which the desired speed is determined v.
(M + m)V 2 /2 = (M + m) gl (1 – cosα).
4. Research of the model and solution of the problem.
The processes that occur when a bullet penetrates a load are no longer purely mechanical. The applied law gives only the lower limit of the estimate - the total energy of the system is conserved, not the mechanical energy - for the right decision problem, we must use the law of conservation of momentum.
« Physics - 10th grade"
Let's calculate the work done by gravity when a body (for example, a stone) falls vertically down.
At the initial moment of time, the body was at a height hx above the Earth's surface, and at the final moment of time - at a height h 2 (Fig. 5.8). Body displacement module |Δ| = h 1 - h 2 .
The directions of the gravity vectors T and displacement Δ coincide. According to the definition of work (see formula (5.2)) we have
A = | T | |Δ|cos0° = mg(h 1 - h 2) = mgh 1 - mgh 2. (5.12)
Now let the body be thrown vertically upward from a point located at a height h 1 above the Earth’s surface, and it reaches a height h 2 (Fig. 5.9). Vectors T and Δ are directed towards opposite sides, and the displacement module |Δ| = h 2 - h 1 . We write the work of gravity as follows:
A = | T | |Δ|cos180° = -mg(h 2 - h 1) = mgh 1 - mgh 2. (5.13)
If the body moves in a straight line so that the direction of movement makes an angle a with the direction of gravity (Fig. 5.10), then the work of gravity is equal to:
A = | T | |Δ|cosα = mg|BC|cosα.
From right triangle BCD it is clear that |BC|cosα = BD = h 1 - h 2 . Hence,
A = mg(h 1 - h 2) = mgh 1 - mgh 2. (5.14)
This expression coincides with expression (5.12).
Formulas (5.12), (5.13), (5.14) make it possible to notice an important regularity. At straight motion body, the work of gravity in each case is equal to the difference between two values of the quantity, depending on the positions of the body, determined by the heights h 1 and h 2 above the Earth's surface.
Moreover, the work done by gravity when moving a body of mass m from one position to another does not depend on the shape of the trajectory along which the body moves. Indeed, if a body moves along the curve BC (Fig. 5.11), then, presenting this curve in the form of a stepped line consisting of vertical and horizontal sections of short length, we will see that in horizontal sections the work of gravity is zero, since the force is perpendicular to the movement , and the amount of work is vertical sections equal to the work that gravity would do when moving a body along a vertical segment of length h 1 - h 2. Thus, the work done by gravity when moving along the curve BC is equal to:
A = mgh 1 - mgh 2.
The work of gravity does not depend on the shape of the trajectory, but depends only on the positions of the initial and end points trajectories.
Let us determine the work A when moving a body along a closed contour, for example along the BCDEB contour (Fig. 5.12). Work A 1 by gravity when moving a body from point B to point D along the trajectory BCD: A 1 = mg(h 2 - h 1), along the trajectory DEB: A 2 = mg(h 1 - h 2).
Then the total work A = A 1 + A 2 = mg(h 2 - h 1) + mg(h 1 - h 2) = 0.
When a body moves along a closed trajectory, the work done by gravity is zero.
So the work of gravity does not depend on the shape of the body's trajectory; it is determined only by the initial and final positions of the body. When a body moves along a closed path, the work done by gravity is zero.
Forces whose work does not depend on the shape of the trajectory of the point of application of the force and is equal to zero along a closed trajectory are called conservative forces.
Gravity is a conservative force.
