Chapter 15. Theorem on the change in kinetic energy.
15.3. Theorem on the change in energy of a kinetic point and a rigid body during translational motion.
15.3.1. How much work is done by the forces acting on a material point if its kinetic energy decreases from 50 to 25 J? (Answer -25)
15.3.2. The free fall of a material point of mass m begins from a state of rest. Neglecting air resistance, determine the distance traveled by the point at the time when it has a speed of 3 m/s. (Answer 0.459)
15.3.3. A material point with mass m = 0.5 kg is thrown from the surface of the Earth with an initial speed v o = 20 m/s and in position M has a speed v= 12 m/s. Determine the work done by gravity when moving a point from position M o to position M (Answer -64)
15.3.4. A material point of mass m is thrown from the Earth's surface at an angle α = 60° to the horizon with initial speed v 0 = 30 m/s. Determine the maximum height h of the rising point. (Answer 34.4)
15.3.5. A body of mass m = 2 kg rises from a push along an inclined plane with an initial speed v o = 2 m/s. Determine the work done by gravity on the path traveled by the body before stopping. (Answer -4)
15.3.6. A material point M of mass m, suspended on a thread of length OM = 0.4 m to a fixed point O, is retracted to an angle α = 90° from the equilibrium position and released without initial velocity. Determine the speed of this point as it passes through the equilibrium position. (Answer 2.80)
15.3.7. The swing cabin is suspended on two rods long l= 0.5 m. Determine the speed of the car when it passes the lower position, if at the initial moment the rods were deflected by an angle φ = 60° and released without initial speed. (Answer 2.21)
15.3.8. A material point M with mass m moves under the influence of gravity along the inner surface of a half-cylinder of radius r = 0.2 m. Determine the speed of the material point at point B of the surface if its speed at point A is zero. (Answer 1.98)
15.3.9. Along wire ABC, located in a vertical plane and bent in the form of arcs of circles of radii r 1, = 1 m, r 2 = 2 m, a ring D of mass m can slide without friction. Determine the speed of the ring at point C if its speed at point A is zero. (Answer 9.90)
15.3.10. A body of mass m = 2 kg moves along a horizontal plane and has been given an initial speed v 0 = 4 m/s. Before stopping, the body traveled a distance of 16 m. Determine the modulus of the sliding friction force between the body and the plane. (Answer 1)
15.3.11. A body with mass m = 100 kg begins to move from rest along a horizontal rough plane under the action of a constant force F. Having covered a path of 5 m, the speed of the body becomes 5 m/s. Determine the modulus of force F if the sliding friction force F tr = 20 N. (Answer 270)
15.3.12.
A hockey player, being at a distance of 10 m from the goal, uses his stick to impart a speed of 8 m/s to the puck lying on the ice. The puck, sliding along the ice surface, flies into the goal at a speed of 7.7 m/s. Determine the coefficient of sliding friction between the puck and the ice surface.
(Answer 2.40 10 -2)
15.3.13. A body of mass m = 1 kg descends down an inclined plane without initial speed. Determine the kinetic energy of the body at the moment of time when it has traveled a distance of 3 m, if the coefficient of sliding friction between the body and the inclined plane f= 0.2. (Answer 9.62)
15.3.14. A load of mass m descends down an inclined plane without initial speed. What speed v will the load have after traveling a distance of 4 m from the start of movement, if the coefficient of sliding friction between the load and the inclined plane is 0.15? (Answer 5.39)
15.3.15.
Spring 2 is attached to slider 1 with mass m = 1 kg. The spring is compressed from its free state by an amount of 0.1 m, after which the load is released without an initial speed. Determine the spring stiffness if the load, having traveled a distance of 0.1 m, acquires a speed of 1 m/s.
(Answer 100)
If a block is pulled with a dynamometer at a constant speed, then the dynamometer shows the modulus of the sliding friction force (F tr). Here the elastic force of the dynamometer spring balances the sliding friction force.
On the other hand, the sliding friction force depends on the force of the normal reaction of the support (N), which arises as a result of the action of the body weight. The greater the weight, the greater the force of the normal reaction. AND the greater the normal reaction force, the greater the friction force. There is a direct proportional relationship between these forces, which can be expressed by the formula:
Here μ is friction coefficient. It shows exactly how the sliding friction force depends on the force of the normal reaction (or, one might say, on the weight of the body), what proportion of it it makes up. The friction coefficient is a dimensionless quantity. For different pairs of surfaces, μ has different values.
For example, wooden objects rub against each other with a coefficient of 0.2 to 0.5 (depending on the type wooden surfaces). This means that if the normal reaction force of the support is 1 N, then during movement the sliding friction force can be a value ranging from 0.2 N to 0.5 N.
From the formula F tr = μN it follows that knowing the forces of friction and normal reaction, you can determine the coefficient of friction for any surfaces:
The strength of the normal ground reaction depends on body weight. It is equal to it in modulus, but opposite in direction. Body weight (P) can be calculated by knowing the body mass. Thus, if we do not take into account the vector nature of quantities, we can write that N = P = mg. Then the friction coefficient is found by the formula:
μ = F tr / (mg)
For example, if it is known that the friction force of a body weighing 5 kg moving on a surface is equal to 12 N, then the friction coefficient can be found: μ = 12 N / (5 kg ∙ 9.8 N/kg) = 12 N / 49 N ≈ 0.245.
2.2.4. Friction force
The friction force acts not only on a moving body, but also on a body at rest, if there are forces that tend to disturb this peace. A friction force also acts on a body that rolls along a support.
Static friction force numerically equal to the component of the force directed along the surface on which the given body is located and tending to move it from its place (Fig. 2.7):
F tr.pok = F x .
Rice. 2.7
When the specified component reaches a certain critical value (F x = F crit), the body begins to move. The critical value of the force, which corresponds to the beginning of movement, is determined by the formula
F x = F crit = µ pok N ,
where µ pok is the coefficient of static friction; N is the modulus of the normal support reaction force (this force is numerically equal to the body weight).
At the moment the movement begins, the static friction force reaches its maximum value:
F tr. pok max = μ pok N .
Sliding friction force is constant and is determined by the product:
F tr.sk = µ sk N ,
where µ sc - sliding friction coefficient; N is the modulus of the normal reaction force of the support.
When solving problems, it is considered that the coefficients of static friction µ pok and sliding friction µ sc are equal to each other:
µ pok = µ sk = µ.
In Fig. Figure 2.8 shows a graph of the dependence of the magnitude of the friction force F tr on the projection of the force F x , tending to move the body, onto the axis directed along the surface of the intended movement.
Rice. 2.8
In order to determine will this body be in rest or starts moving under the influence of an applied force of a certain magnitude and direction, it is necessary:
F crit = µN,
where µ is the friction coefficient; N is the modulus of the normal reaction force of the support;
3) compare the values of F crit and F x:
- if F x > F crit, then the body moves under the action of the applied force; in this case, the sliding friction force is calculated as
F tr.sk = µN ;
- if F x< F крит, то тело покоится под действием приложенной силы; в этом случае сила трения покоя рассчитывается как
F tr.pok = F x .
Module rolling friction forces F roller swing is proportional to the rolling friction coefficient µ roll, the modulus of the normal reaction force of the support N and is inversely proportional to the radius R of the rolling body:
F tr. quality = μ quality N R .
Example 13. A force of 25 N directed along the surface is applied to a body with a mass of 6.0 kg lying on a horizontal surface. Find the friction force if the friction coefficient is 0.5.
Solution. Let us estimate the magnitude of the force that can cause the movement of a body using the formula
Fcr = µN,
where µ is the friction coefficient; N is the modulus of the normal reaction force of the support, numerically equal to the body weight (P = mg).
The magnitude of the critical force sufficient to start the movement of the body is
F cr = μ m g = 0.5 ⋅ 6.0 ⋅ 10 = 30 N.
The projection of the force applied to the body in the horizontal direction onto the axis of the expected motion Ox (see figure) is equal to
F x = F = 25 N.
Fx< F кр,
those. the magnitude of the force applied to the body is less than the magnitude of the force capable of causing its movement. Therefore, the body is at rest.
The desired friction force - the static friction force - is equal to the external horizontal force tending to disrupt this peace:
F tr.pok = F x = 25 N.
Example 14. The body is on an inclined plane with a base angle of 30°. Calculate the friction force if the friction coefficient is 0.5 3. Body weight is 3.0 kg.
Solution. The arrow in the figure shows the direction of the expected movement.
Let's find out whether the body will remain at rest or begin to move. To do this, let’s calculate the magnitude of the critical force that can cause movement, i.e.
Fcr = µN,
where µ is the friction coefficient; N = mg cos α is the magnitude of the normal reaction force of the inclined plane.
The calculation gives the value of the indicated force:
F cr = μ m g cos 30 ° = 0.5 3 ⋅ 3.0 ⋅ 10 ⋅ 3 2 = 22.5 N.
The body tends to be brought out of a state of rest by the projection of gravity onto the Ox axis, the magnitude of which is
F x = mg sin 30° = 15 N.
Thus, there is an inequality
Fx< F кр,
those. the projection of the force tending to cause the movement of the body is less than the magnitude of the force capable of doing this. Consequently, the body maintains a state of rest.
The required force - the force of static friction - is equal to
F tr = F x = 15 N.
Example 15. The washer is located on the inner surface of the hemisphere at a height of 10 cm from the bottom point. The radius of the hemisphere is 50 cm. Calculate the coefficient of friction of the washer on the sphere if it is known that the indicated height is the maximum possible.
Solution. Let us illustrate the problem condition with a drawing.
The puck, according to the conditions of the problem, is at the maximum possible height. Consequently, the static friction force acting on the washer has a maximum value that coincides with the projection of gravity on the Ox axis:
F tr. so far max = F x ,
where F x = mg cos α is the modulus of the projection of gravity onto the Ox axis; m is the mass of the washer; g - free fall acceleration module; α is the angle shown in the figure.
The maximum static friction force coincides with the sliding friction force:
F tr. until max = F tr. sk,
where F tr.sk = µN - modulus of sliding friction force; N = mg sin α is the magnitude of the normal reaction force of the surface of the hemisphere; µ - friction coefficient.
Let us determine the friction coefficient by writing the indicated equality explicitly:
mg cos α = µmg sin α.
It follows that the desired friction coefficient is determined by the tangent of the angle α:
We determine the indicated angle from an additional construction:
tg α = R − h 2 h R − h 2 ,
where h is the maximum height at which the washer can be located; R is the radius of the hemisphere.
The calculation gives the tangent value:
tan α = 0.5 − 0.1 2 ⋅ 0.1 ⋅ 0.5 − (0.1) 2 = 4 3
and allows you to calculate the required friction coefficient.
The friction force () is the force that arises during the relative motion of bodies. It has been empirically established that the force of sliding friction depends on the force of mutual pressure of the bodies (support reaction) (N), the materials of the surfaces of the rubbing bodies, and the speeds of relative motion.
DEFINITION
The physical quantity that characterizes rubbing surfaces is called friction coefficient. Most often, the friction coefficient is denoted by the letters k or.
In general, the friction coefficient depends on the speed of movement of bodies relative to each other. It should be noted that the dependence is usually not taken into account and the sliding friction coefficient is considered constant. In most cases, the friction force
The sliding friction coefficient is a dimensionless quantity. The coefficient of friction depends on: the quality of surface treatment, rubbing bodies, the presence of dirt on them, the speed of movement of bodies relative to each other, etc. The friction coefficient is determined empirically (experimentally).
The friction coefficient, which corresponds to the maximum static friction force, is in most cases greater than the sliding friction coefficient.
For more pairs of materials, the friction coefficient is no more than unity and lies within
The value of the friction coefficient of any pair of bodies between which the friction force is considered is influenced by pressure, degree of contamination, surface area of the bodies, and other things that are usually not taken into account. Therefore, the values of the friction force coefficients that are indicated in the reference tables completely coincide with reality only under the conditions under which they were obtained. Consequently, the values of the coefficients of friction forces cannot be considered unchanged for the same pair of rubbing bodies. Thus, thorn coefficients are distinguished for dry surfaces and lubricated surfaces. For example, the sliding coefficient for a body made of bronze and a body made of cast iron, if the surfaces of the materials are dry, is equal to For the same pair of materials, the sliding coefficient in the presence of lubrication
Examples of problem solving
EXAMPLE 1
| Exercise | A thin metal chain lies on a horizontal table (Fig. 1). Its length is equal to , mass . The end of the chain hangs over the edge of the table. If the length of the hanging part of the chain is a fraction of the length of the entire chain, it begins to slide down the table. What is the coefficient of friction between the chain and the table if the chain is considered uniform in length? |
| Solution | The chain moves under the influence of gravity. Let the force of gravity acting per unit length of the chain be equal to . In this case, at the moment the sliding begins, the force of gravity acting on the overhanging part will be: Before sliding begins, this force is balanced by the frictional force that acts on the part of the chain that lies on the table: Since the forces are balanced, we can write (): |
| Answer |
EXAMPLE 2
| Exercise | What is the coefficient of friction of a body on an inclined plane if the angle of inclination of the plane is equal and its length is equal to . The body moved along the plane with constant acceleration during time t. |
| Solution | In accordance with Newton's second law, the resultant of forces applied to a body moving with acceleration is equal to: In projections on the X and Y axes of equation (2.1), we obtain: |
Physics workshop
Problem No. 3
Determination of sliding friction coefficient
In preparation for this task, you should familiarize yourself with the theory in the textbooks:
1. Chapter 2, I.V. Savelyev “Course of General Physics”, vol. 1, M., “Science”.
2. § 1 and 2. PC. Kashkarov, A.V. Zoteev, A.N. Nevzorov, A.A. Sklyankin “Problems for the course”general physics with solutions. « Mechanics. Electricity and magnetism » , M., ed. Moscow State University.
Goal of the work
Experimentally test the laws of kinematics and dynamics using the example of translational motion of a rigid body in the presence of dry friction. Get acquainted with the method of determining the coefficient of sliding friction - tribometry. Based on experimental data, calculate the sliding friction coefficient.
2. Experimental equipment, instruments and accessories
L
laboratory stand (Fig. 3.1) includes an inclined guide bench (1) with a measuring ruler attached to it, a movable block (2) (2 pcs.), optical sensors (3) (3 pcs.), a protractor for measuring the angle of inclination of the guide bench and a module for collecting signals from optical sensors (4).
Equipment and accessories include a computer with the necessary software and a hub for connecting the signal acquisition module to a computer.
3. Theoretical part
A. General provisions
When analyzing the motion of bodies using Newton's laws, one has to deal with the following types of forces:
Gravity is a manifestation of the gravitational interaction of bodies;
Tension force of threads, springs, reaction of supports and suspensions, etc. (“reaction forces of bonds”) – a manifestation of elastic forces that arise during the deformation of bodies;
Friction force . Distinguish between forces dry and viscous friction. Dry friction occurs when a solid body can move on the surface of another solid body.
In conditions when forces act on a body in contact with a certain surface, but it does not move relative to this surface, the latter acts on the body static friction force . Its value is found from the condition of the absence of relative motion:
(3.1),
Where 
– forces applied to the body, with the exception of 
. Those. while the body is at rest, the static friction force is exactly equal in magnitude and opposite in direction to the tangential component of the resultant forces 
. Maximum friction force n 
okoya equals 
, Where N
– normal (i.e. perpendicular to the surfaces) component ground reaction forces *) ,
– coefficient of sliding friction. The friction coefficient depends on the material and condition of the surfaces of the contacting bodies. For rough surfaces the coefficient of friction is greater than for polished ones. In Fig. 3.2 shows how the force of dry friction changes with increasing force magnitude
F
. Sloping section of the graph ( F tr N) corresponds to a body at rest ( F tr pok = F
), and horizontal – sliding.
. (3.2)
* By their nature, dry friction forces are caused by the electromagnetic interaction of molecules of the surface layers of contacting solids. The independence of the friction force from speed is observed only at not very high speeds, not for all bodies and not for all surface processing qualities.
The sliding friction force is always directed opposite to the velocity vector of the body. This corresponds to the vector representation of the law for the sliding friction force, established experimentally by the French physicists C. Coulomb and G. Amonton:
. (3.3)
Here 
– speed of relative motion of bodies, v– its module.
When bodies move in liquid or gaseous media, viscous friction force . At low speeds it is proportional to the speed of movement of the body relative to the medium:
, (3.4)
Where r – coefficient of viscous friction (depends on the size and shape of the body, on the viscous properties of the medium).
The system of methods for measuring forces, friction coefficients and wear resistance of rubbing bodies constitutes the content special section mechanics – tribometry. In this work, to experimentally determine the sliding friction coefficient a tribometer is used in the form of an inclined plane with an adjustable angle of inclination and a system of optical sensors to record the kinematic characteristics of a body sliding from it.
B. Derivation of the “calculation formula”
B 
The handle located on the inclined plane of the guide bench of the laboratory bench (Fig. 3.1) experiences the action of two forces: gravity 
and the support reaction force from the wedge. The latter, as usual, is convenient to immediately imagine in the form of two components - friction force 
along the surface and the “normal” component (i.e. perpendicular to the surface) – 
(see Fig. 3.3). In general, the friction force can be directed both up and down along an inclined plane. However, we will be interested in the case where the block is either sliding or is on the verge of sliding down an inclined plane. Then the friction force is directed obliquely downwards.
We will assume that the stand is stationary relative to the inertial frame of reference associated with the Earth. Then, until the block slides, the sum of the forces acting on it is zero. Convenient axle ABOUTX And ABOUTY The coordinate systems of the inertial reference system we choose should be placed along the inclined plane and perpendicular to it, respectively (see Fig. 3.3). The equilibrium conditions for a block at rest on an inclined plane have the form:
0 = N – mgcos . (3.5)
0 = mgsin – F tr . (3.6)
While the angle of inclination of the guide is small, the component of gravity along it (“rolling force”) is balanced by the force static friction (!). As the angle increases it also grows (according to the “law of sine”). However, its growth is not unlimited. Its maximum value, as we know, is equal to
=
N.
(3.7)
This determines the maximum value of the angle at which the block does not slide off the inclined plane. Joint decision equations (3.5) – (3.7) leads to the condition:
. (3.8)
In other words, friction coefficient equal to the tangent of the angle of inclination of the plane to the horizon at which sliding begins bodies from an inclined plane. This is the basis of the operating principle of one of the possible options tribometers.
However, it is quite difficult to establish with sufficient accuracy the limiting angle at which a body begins to slide off an inclined plane (“static method”). Therefore, in this experimental work, a dynamic method is used to determine the sliding friction coefficient during translational motion of a rigid body (bar) along an inclined plane with acceleration.
When a block slides down an inclined plane, the equation of motion (Newton’s second law) in projections onto the coordinate axes will look like this:
ma= mg sin – F tr , (3.9)
0 = Nmg cos . (3.10)
The sliding friction force is equal to
F tr = N . (3.11)
These dynamics equations allow you to find the acceleration of the body:
a= (sin – cos )g. (3.12)
The coordinate of a body sliding down an inclined plane changes according to the law uniformly accelerated motion:
. (3.13)
Optical sensors placed at fixed distances along the path of movement of the block make it possible to measure the time it takes the body to travel through the corresponding sections of the path. Using equality (3.13), by numerical approximation of experimental data, we can find the acceleration value a.
Based on the value of the calculated acceleration, using equality (3.12), one can obtain a “calculation formula” for determining the friction coefficient :
(3.14)
Thus, to experimentally determine the friction coefficient, it is necessary to measure two quantities: the angle of inclination of the plane and body acceleration A.
Description of the laboratory setup
D 
Rice. 3.4
Wooden block 1 (Fig. 3.4) with a sight strip (2) of length glued to it ℓ , slides along an inclined plane, crossing the optical axes of sensors (3), which record the moments of the beginning and end of the overlap of their optical axes with a block sliding along the inclined plane. The leading edge of the pulse of the optical axis of the sensor is associated with the beginning of the overlap of the optical axis by the target bar, and the trailing edge is associated with the completion of the overlap of the optical axis by the bar. During this time, the block moves a distance ℓ . Thus, when a bar sequentially intercepts the optical axes of three sensors, the times of passage of 6 coordinate marks on the axis are recorded OH(see Fig. 3.5): x 1 , x 1 +ℓ , x 2 , x 2 +ℓ , x 3 , x 3 +ℓ . Experimentally measured values of their transit times t 1 ,t 2 ,t 3 ,t 4 ,t 5 ,t 6 serve as the basis for approximating the quadratic dependence curve (3.13). The approximation program must include the coordinate values of these points x 1 , x 1 +ℓ , x 2 , x 2 +ℓ , x 3 , x 3 +ℓ , which are entered into Table 1 after fixing the positions of 3 optical sensors.
Work procedure
Installation options:
Length of the bar sighting bar: ℓ = (110 1) mm;
Angles of inclination of the bench guide for bars No. 1 and No. 2:
α 1 = (24 ± 1) hail;
α 2 = (27 ± 1) hail.
Table 1
Coordinate 1st sensor x 1 , mm | x 1 +ℓ , | Coordinate 2nd sensor x 2 , mm | x 2 +ℓ , | Coordinate 3rd sensor x 3 , mm | x 3 +ℓ , |
Exercise 1 (bar No. 1)
1. Assemble the laboratory setup by installing the guide bench at an angle α 1 = 24 (controlled using a protractor) and placing 3 optical sensors along the path of the block along the bench guide.
2. Place block No. 1 on the inclined guide and hold it in the upper, initial position.
Start measurements by pressing the (Ctrl+S) button (start measurements for the selected sensors) and immediately, immediately after the start, release the block, after which it will begin to slide along the inclined plane from the top position.
3. After the block passes the entire inclined plane, stop the measurements by pressing the (Ctrl+T) button (stop measurements). Three pulses will be visible on the screen, showing the moments of overlap of the optical axes of 3 sensors when a wooden block slides along an inclined plane (Fig. 3.6) (the numbers are relative).
R 
is. 3.6
4. Process the received data in accordance with the scenario:
t,With | x,m |
|
the right column of the table, marked " x, m", must be filled in manually. If three sensors are installed at 15 cm, 40 cm and 65 cm accordingly (the data is taken from table 1), then after entering all six values of the sensor coordinates, the table on the screen will look like this:
t,With | x,m |
|
the figure in the central column of the table (under the designation “A”) is equal to twice the coefficient of the quadratic power in equation (3.13), i.e. 
, so in this case the acceleration value will be equal to a 1 = 2A = 0.13×2 = 0.26 m/With 2. Record this value in table 2.
5. Repeat the experiment according to paragraphs. 2-4 four more times. Record all results in Table 2.
6. Set the guide bench at an angle α 2 = 27 , by placing three optical sensors in the path of the block moving along the bench guide. Repeat the entire experiment according to paragraphs. 2–4. Record all results in Table 3.
Table 2, block No. 1 ( α 1 = 24)
| Table 3, block No. 1( α 2 = 27)
|
After the tables, leave space for recording the calculated results (about half a page).
Exercise 2 (bar No. 2)
1. Take block No. 2 with a different material for the supporting sliding surface and repeat the entire experiment according to paragraphs. 1–6. Record all results in tables 4 and 5, respectively.
Table 4, block No. 2 ( α 1 = 24)
| Table 5, block No. 2 ( α 2 = 27)
|
After the tables, leave space for the calculated results (about half a page).
6. Processing of measurement results
Using the results obtained and the calculated relationship (3.14), find the average value of the friction coefficient I>μ> for each block and experimental conditions (angle of inclination of the plane):
…
Record partial deviations in tables 2–4. Find the measurement error for each case
For bar No. 1:
1 > =…; 2 > = …;
For bar No. 2:
3 > = …; 4 > = …
2. Assess the experimental error (measurement error + method error).
Measurement error (average of partial deviation modules):
= ...
Δ µ 1 change = …;Δ µ 2 change = …;
Δ µ 3 change = …;Δ µ 4 change = …
Method error:
/B> a 1 > = … m/s 2 ;Δ a 1 = … m/s 2
ε µ = … Δ µ 1 meth = ε µ · 1 > = …
Δ µ 1 = …
/B> a 2 > = … m/s 2 ;Δ a 2 = … m/s 2
ε µ = ... Δ µ 2 meth = ε µ · 2 > = …
Δ µ 2 = …
/B> a 3 > = … m/s 2 ;Δ a 3 = … m/s 2
ε µ = ... Δ µ 3 meth = ε µ · 3 > = ...
Δ µ 3 =
/B> a 4 > = … m/s 2 ;Δ a 4 = … m/s 2
ε µ = ... Δ µ 4 meth = ε µ · 4 > =
Δ µ 4 = ...
Write down the result of the experimental determination of the friction coefficient μ for bar No. 1 and for bar No. 2 in standard form:
7. Test questions
What is friction force?
What types of friction forces do you know?
What is static friction force? What is the static friction force?
Draw graphs of the dependence of the dry friction force on the tangent to the surface of the support and the resulting component of the remaining forces acting on the body.
What does the coefficient of sliding friction depend on?
How can you experimentally determine the coefficient of sliding friction from the conditions of equilibrium of a body on an inclined plane?
How is the sliding friction coefficient experimentally determined in this work?
What is a laboratory bench?
Tell us about the procedure for performing the work and taking measurements.
How to estimate the error of indirect measurement of the sliding friction coefficient?
8. Safety instructions
Before performing work, receive instructions from a laboratory assistant.
Follow the general safety rules for working in the Physics laboratory.
9. Applications
Appendix 1. Estimation of measurement error using coefficients Lesson
Number for the purpose of collecting statistical data): definitioncoefficientfrictionslip bodies on the surface used (use... us tasks? – The acceleration of the body must be zero. – At what value coefficientfriction ...
Rectilinear movement" 1 3 Solving graphical tasks 1 4 Solution tasks
