Instruction

Case 1. The formula for sliding: Ftr = mN, where m is the coefficient of sliding friction, N is the reaction force of the support, N. For a body sliding along a horizontal plane, N = G = mg, where G is the weight of the body, N; m – body weight, kg; g is the free fall acceleration, m/s2. The values ​​of the dimensionless coefficient m for a given pair of materials are given in the reference. Knowing the mass of the body and a couple of materials. sliding relative to each other, find the force of friction.

Case 2. Consider a body sliding on a horizontal surface and moving with uniform acceleration. Four forces act on it: the force that sets the body in motion, the force of gravity, the reaction force of the support, the force of sliding friction. Since the surface is horizontal, the reaction force of the support and the force of gravity are directed along one straight line and balance each other. The displacement describes the equation: Fdv - Ftr = ma; where Fdv is the modulus of force that sets the body in motion, N; Ftr is the friction force modulus, N; m – body weight, kg; a is acceleration, m/s2. Knowing the values ​​​​of the mass, acceleration of the body and the force acting on it, find the force of friction. If these values ​​are not set directly, see if there is data in the condition from which to find these values.

Example of problem 1: a 5 kg bar lying on the surface is subjected to a force of 10 N. As a result, the bar moves with uniform acceleration and passes 10 for 10. Find the force of sliding friction.

The equation for the movement of the bar: Fdv - Ftr \u003d ma. Body path for uniformly accelerated motion is given by the equality: S = 1/2at^2. From here you can determine the acceleration: a = 2S/t^2. Substitute these conditions: a \u003d 2 * 10 / 10 ^ 2 \u003d 0.2 m / s2. Now find the resultant of the two forces: ma = 5 * 0.2 = 1 N. Calculate the friction force: Ftr = 10-1 = 9 N.

Case 3. If a body on a horizontal surface is at rest or moves uniformly, according to Newton's second law, the forces are in equilibrium: Ftr = Fdv.

Problem 2 example: a 1 kg bar on a flat surface is told , as a result of which it travels 10 meters in 5 seconds and stops. Determine the force of sliding friction.

As in the first example, the sliding of the bar is affected by the force of motion and the force of friction. As a result of this action, the body stops, i.e. balance comes. The equation of motion of the bar: Ftr = Fdv. Or: N*m = ma. The block slides with uniform acceleration. Calculate its acceleration similarly to problem 1: a = 2S/t^2. Substitute the values ​​of the quantities from the condition: a \u003d 2 * 10 / 5 ^ 2 \u003d 0.8 m / s2. Now find the friction force: Ftr \u003d ma \u003d 0.8 * 1 \u003d 0.8 N.

Case 4. Three forces act on a body spontaneously sliding along an inclined plane: gravity (G), support reaction force (N) and friction force (Ftr). The force of gravity can be written as follows: G = mg, N, where m is the body weight, kg; g is the free fall acceleration, m/s2. Since these forces are not directed along a single straight line, write the equation of motion in vector form.

By adding the forces N and mg according to the parallelogram rule, you get the resultant force F'. The following conclusions can be drawn from the figure: N = mg*cosα; F' = mg*sinα. Where α is the angle of inclination of the plane. The friction force can be written by the formula: Ftr = m*N = m*mg*cosα. The equation for motion takes the form: F’-Ftr = ma. Or: Ftr = mg*sinα-ma.

Case 5. If an additional force F is applied to the body, directed along an inclined plane, then the friction force will be expressed: Ftr = mg * sinα + F-ma, if the direction of movement and force F are the same. Or: Ftr \u003d mg * sinα-F-ma, if the force F opposes the movement.

Problem 3 Example: A 1 kg block slid down the top of an inclined plane in 5 seconds after traveling a distance of 10 meters. Determine the force of friction if the angle of inclination of the plane is 45o. Consider also the case where the block was subjected to an additional force of 2 N applied along the angle of inclination in the direction of travel.

Find the acceleration of the body in the same way as in examples 1 and 2: a = 2*10/5^2 = 0.8 m/s2. Calculate the friction force in the first case: Ftr \u003d 1 * 9.8 * sin (45o) -1 * 0.8 \u003d 7.53 N. Determine the friction force in the second case: Ftr \u003d 1 * 9.8 * sin (45o) +2-1*0.8= 9.53 N.

Case 6. A body moves uniformly along an inclined surface. So, according to Newton's second law, the system is in equilibrium. If the sliding is spontaneous, the motion of the body obeys the equation: mg*sinα = Ftr.

If an additional force (F) is applied to the body, which prevents uniformly accelerated movement, the expression for motion has the form: mg*sinα–Ftr-F = 0. From here, find the friction force: Ftr = mg*sinα-F.

Sources:

• slip formula

The coefficient of friction is a combination of the characteristics of two bodies that are in contact with each other. There are several types of friction: static friction, sliding friction and rolling friction. Resting friction is the friction of a body that was at rest and was set in motion. Sliding friction occurs when the body moves, this friction is less than static friction. Rolling friction occurs when a body rolls on a surface. Friction is designated depending on the type, as follows: μsk - sliding friction, μ - static friction, μroll - rolling friction.

Instruction

When determining the coefficient of friction during the experiment, the body is placed on a plane at an inclination and the angle of inclination is calculated. At the same time, take into account that when determining the coefficient of static friction, the given body moves, and when determining the coefficient of sliding friction, it moves at a constant speed.

The coefficient of friction can also be calculated during the experiment. It is necessary to place the object on an inclined plane and calculate the angle of inclination. Thus, the coefficient of friction is determined by the formula: μ=tg(α), where μ is the friction force, α is the angle of inclination of the plane.

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At relative motion friction between two bodies. It can also occur when moving in a gaseous or liquid medium. Friction can both interfere with and contribute to normal movement. As a result of this phenomenon, a force acts on the interacting bodies friction.

Instruction

Most general case considers the force when one of the bodies is fixed and at rest, and the other slides on its surface. From the side of the body on which the moving body slides, the reaction force of the support acts on the latter, directed perpendicular to the plane of sliding. This force is represented by the letter N. The body can also be at rest relative to the fixed body. Then the friction force acting on it Ffr

In the case of body motion relative to the surface of a fixed body, the sliding friction force becomes equal to the product of the friction coefficient and the reaction force of the support: Ftr = ?N.

Let now a constant force F>Ftr = ?N, parallel to the surface of the contacting bodies, acts on the body. When the body slides, the resulting component of the force in the horizontal direction will be equal to F-Ftr. Then, according to Newton's second law, the acceleration of the body will be associated with the resulting force according to the formula: a = (F-Ftr)/m. Hence, Ftr = F-ma. The acceleration of the body can be found from kinematic considerations.

The often considered special case of the friction force manifests itself when a body slides off a fixed inclined plane. Let be? - the angle of inclination of the plane and let the body slide evenly, that is, without acceleration. Then the equations of motion of the body will look like this: N = mg*cos?, mg*sin? = Ftr = ?N. Then, from the first equation of motion, the friction force can be expressed as Ftr = ?mg*cos?. If the body moves along an inclined plane with acceleration a, then the second equation of motion will look like: mg*sin?-Ftr = ma. Then Ftr = mg*sin?-ma.

Related videos

If the force directed parallel to the surface on which the body stands exceeds the static friction force, then motion will begin. It will continue until the driving force exceeds the sliding friction force, which depends on the coefficient of friction. You can calculate this coefficient yourself.

You will need

• Dynamometer, scales, protractor or goniometer

Instruction

Find the weight of the body in kilograms and place it on a flat surface. Attach a dynamometer to it, and start moving the body. Do this in such a way that the dynamometer readings stabilize while maintaining a constant speed. In this case, the traction force measured by the dynamometer will be equal, on the one hand, to the traction force shown by the dynamometer, and on the other hand, to the force multiplied by the slip.

The measurements made will allow you to find this coefficient from the equation. To do this, divide the traction force by the mass of the body and the number 9.81 (gravitational acceleration) μ=F/(m g). The coefficient obtained will be the same for all surfaces of the same type as those on which the measurement was made. For example, if the body from moved along a wooden board, then this result will be valid for all wooden bodies sliding along the tree, taking into account the quality of its processing (if the surfaces are rough, the value of the sliding friction coefficient will change).

You can measure the coefficient of sliding friction in another way. To do this, place the body on a plane that can change its angle relative to the horizon. It can be an ordinary board. Then begin to gently lift it by one edge. At the moment when the body begins to move, rolling down in a plane like a sled down a hill, find the angle of its slope relative to the horizon. It is important that the body does not move with acceleration. In this case, the measured angle will be extremely small, at which the body will begin to move under the action of gravity. The coefficient of sliding friction will be equal to the tangent of this angle μ=tg(α).

Force of normal reaction- the force acting on the body from the side of the support (or suspension). When the bodies come into contact, the reaction force vector is directed perpendicular to the contact surface. The following formula is used for the calculation:

$|\backslash vec\; N|=\; mg\; \backslash cos\; \backslash theta,$

where $|\backslash vecN|$ is the modulus of the normal reaction force vector, $m$- body mass, $g$- acceleration of gravity , $\backslash theta$- the angle between the support plane and the horizontal plane.

According to Newton's third law, the modulus of the normal reaction force $|\backslash vecN|$ equal to the modulus of body weight $|\backslash vec\; P|$, but their vectors are collinear and oppositely directed:

$\backslash vec\; N=\; -\backslash vec\; P.$

It follows from the Amonton-Coulomb law that the following relation is true for the modulus of the normal reaction force vector:

$|\backslash vec\; N|=\; \backslash frac(|\backslash vec\; F|)(k),$

where $\backslash vec\; F$- sliding friction force, and $k$- coefficient of friction.

Since the static friction force is calculated by the formula

$|\backslash vec\; f|=\; mg\; \backslash sin\; \backslash theta,$

then we can experimentally find such an angle value $\backslash theta$, at which the static friction force will be equal to the sliding friction force:

$mg\; \backslash sin\; \backslash theta\; =\; k\; mg\; \backslash cos\; \backslash theta.$

From here we express the coefficient of friction:

$k\; =\; \backslash mathrm(tg)\backslash \; \backslash theta.$

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An excerpt characterizing the Force of a normal reaction

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No matter how strange the historical descriptions of how some king or emperor, having quarreled with another emperor or king, gathered an army, fought with the army of the enemy, won a victory, killed three, five, ten thousand people and, as a result, conquered the state and the whole people in several million; no matter how incomprehensible why the defeat of one army, one hundredth of all the forces of the people, forced the people to submit, - all the facts of history (as far as we know) confirm the justice of the fact that greater or lesser successes of the army of one people against the army of another people are the causes or, according to at least essential signs of an increase or decrease in the strength of the peoples. The army won, and immediately the rights of the victorious people increased to the detriment of the defeated. The army has suffered a defeat, and immediately, according to the degree of defeat, the people are deprived of their rights, and with the complete defeat of their army, they completely submit.
So it has been (according to history) from ancient times to the present. All the wars of Napoleon serve as confirmation of this rule. According to the degree of defeat of the Austrian troops - Austria is deprived of its rights, and the rights and forces of France increase. The victory of the French at Jena and Auerstet destroys the independent existence of Prussia.

The force acting on the body from the side of the support (or suspension) is called the reaction force of the support. When the bodies come into contact, the reaction force of the support is directed perpendicular to the contact surface. If the body lies on a horizontal fixed table, the reaction force of the support is directed vertically upwards and balances the force of gravity:

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It is necessary to know the point of application and the direction of each force. It is important to be able to determine exactly what forces act on the body and in what direction. Force is denoted as , measured in Newtons. In order to distinguish between forces, they are designated as follows

Below are the main forces acting in nature. It is impossible to invent non-existent forces when solving problems!

There are many forces in nature. Here we consider the forces that are considered in the school physics course when studying dynamics. Other forces are also mentioned, which will be discussed in other sections.

The force of gravity

Every body on the planet is affected by the Earth's gravity. The force with which the Earth attracts each body is determined by the formula

The point of application is at the center of gravity of the body. The force of gravity always pointing vertically down.

Friction force

Let's get acquainted with the force of friction. This force arises when bodies move and two surfaces come into contact. The force arises as a result of the fact that the surfaces, when viewed under a microscope, are not smooth as they seem. The friction force is determined by the formula:

A force is applied at the point of contact between two surfaces. Directed in the direction opposite to the movement.

Support reaction force

Imagine a very heavy object lying on a table. The table bends under the weight of the object. But according to Newton's third law, the table acts on the object with exactly the same force as the object on the table. The force is directed opposite to the force with which the object presses on the table. That is up. This force is called the support reaction. The name of the force "speaks" react support. This force arises whenever there is an impact on the support. The nature of its occurrence at the molecular level. The object, as it were, deformed the usual position and connections of the molecules (inside the table), they, in turn, tend to return to their original state, "resist".

Absolutely any body, even a very light one (for example, a pencil lying on a table), deforms the support at the micro level. Therefore, a support reaction occurs.

There is no special formula for finding this force. They designate it with the letter, but this force is just a separate type of elastic force, so it can also be denoted as

The force is applied at the point of contact of the object with the support. Directed perpendicular to the support.

Since the body is represented as a material point, the force can be depicted from the center

Elastic force

This force arises as a result of deformation (changes in the initial state of matter). For example, when we stretch a spring, we increase the distance between the molecules of the spring material. When we compress the spring, we decrease it. When we twist or shift. In all these examples, a force arises that prevents deformation - the elastic force.

Hooke's Law

The elastic force is directed opposite to the deformation.

Since the body is represented as a material point, the force can be depicted from the center

When connected in series, for example, springs, the stiffness is calculated by the formula

When connected in parallel, the stiffness

Sample stiffness. Young's modulus.

Young's modulus characterizes the elastic properties of a substance. This is a constant value that depends only on the material, its physical state. Characterizes the ability of a material to resist tensile or compressive deformation. The value of Young's modulus is tabular.

Learn more about the properties of solids.

Body weight

Body weight is the force with which an object acts on a support. You say it's gravity! The confusion occurs in the following: indeed, often the weight of the body is equal to the force of gravity, but these forces are completely different. Gravity is the force that results from interaction with the Earth. Weight is the result of interaction with the support. The force of gravity is applied at the center of gravity of the object, while the weight is the force that is applied to the support (not to the object)!

There is no formula for determining weight. This force is denoted by the letter .

The support reaction force or elastic force arises in response to the impact of an object on a suspension or support, therefore the body weight is always numerically the same as the elastic force, but has the opposite direction.

The reaction force of the support and the weight are forces of the same nature, according to Newton's 3rd law they are equal and oppositely directed. Weight is a force that acts on a support, not on a body. The force of gravity acts on the body.

Body weight may not be equal to gravity. It can be either more or less, or it can be such that the weight is zero. This state is called weightlessness. Weightlessness is a state when an object does not interact with a support, for example, the state of flight: there is gravity, but the weight is zero!

It is possible to determine the direction of acceleration if you determine where the resultant force is directed

Note that weight is a force, measured in Newtons. How to correctly answer the question: "How much do you weigh"? We answer 50 kg, naming not weight, but our mass! In this example, our weight is equal to gravity, which is approximately 500N!

Overload- the ratio of weight to gravity

Strength of Archimedes

Force arises as a result of the interaction of a body with a liquid (gas), when it is immersed in a liquid (or gas). This force pushes the body out of the water (gas). Therefore, it is directed vertically upwards (pushes). Determined by the formula:

In the air, we neglect the force of Archimedes.

If the Archimedes force is equal to the force of gravity, the body floats. If the Archimedes force is greater, then it rises to the surface of the liquid, if it is less, it sinks.

electrical forces

There are forces of electrical origin. Occur in the presence of an electric charge. These forces, such as the Coulomb force, Ampère force, Lorentz force, are discussed in detail in the Electricity section.

Schematic designation of the forces acting on the body

Often the body is modeled by a material point. Therefore, in the diagrams, various points of application are transferred to one point - to the center, and the body is schematically depicted as a circle or rectangle.

In order to correctly designate the forces, it is necessary to list all the bodies with which the body under study interacts. Determine what happens as a result of interaction with each: friction, deformation, attraction, or maybe repulsion. Determine the type of force, correctly indicate the direction. Attention! The number of forces will coincide with the number of bodies with which the interaction takes place.

The main thing to remember

1) Forces and their nature;
2) Direction of forces;
3) Be able to identify the acting forces

Distinguish between external (dry) and internal (viscous) friction. External friction occurs between solid surfaces in contact, internal friction occurs between layers of liquid or gas during their relative motion. There are three types of external friction: static friction, sliding friction and rolling friction.

Rolling friction is determined by the formula

The resistance force arises when a body moves in a liquid or gas. The magnitude of the resistance force depends on the size and shape of the body, the speed of its movement and the properties of the liquid or gas. At low speeds, the resistance force is proportional to the speed of the body

At high speeds it is proportional to the square of the speed

Consider the mutual attraction of an object and the Earth. Between them, according to the law of gravity, a force arises

Now let's compare the law of gravity and the force of gravity

The value of free fall acceleration depends on the mass of the Earth and its radius! Thus, it is possible to calculate with what acceleration objects on the Moon or on any other planet will fall, using the mass and radius of that planet.

The distance from the center of the Earth to the poles is less than to the equator. Therefore, the acceleration of free fall at the equator is slightly less than at the poles. At the same time, it should be noted that the main reason for the dependence of the acceleration of free fall on the latitude of the area is the fact that the Earth rotates around its axis.

When moving away from the surface of the Earth, the force of gravity and the acceleration of free fall change inversely with the square of the distance to the center of the Earth.

Let's put a stone on a horizontal table top, standing on the ground (Fig. 104). Since the acceleration of a stone relative to the Earth is equal to a bullet, then according to Newton's second law, the sum of the forces acting on it is zero. Consequently, the action of the gravity force m · g on the stone must be compensated by some other forces. It is clear that under the action of the stone the table top is deformed. Therefore, from the side of the table, an elastic force acts on the stone. If we assume that the stone interacts only with the Earth and the table top, then the elastic force must balance the force of gravity: F control = -m · g. This elastic force is called support reaction force and are denoted by the Latin letter N. Since the acceleration of free fall is directed vertically downwards, the force N is directed vertically upwards - perpendicular to the surface of the table top.

Since the table top acts on the stone, then, according to Newton's third law, the stone also acts on the table top with the force P = -N (Fig. 105). This force is called weighing.

The weight of a body is the force with which this body acts on a suspension or support, being in a stationary state relative to the suspension or support.

It is clear that in the considered case the weight of the stone is equal to the force of gravity: P = m · g. This will be true for any body resting on a suspension (support) relative to the Earth (Fig. 106). Obviously, in this case, the attachment point of the suspension (or support) is stationary relative to the Earth.

For a body resting on a suspension (support) that is motionless relative to the Earth, the weight of the body is equal to the force of gravity.

The weight of the body will also be equal to the force of gravity acting on the body if the body and the suspension (support) move uniformly in a straight line relative to the Earth.

If the body and the suspension (support) move relative to the Earth with acceleration so that the body remains stationary relative to the suspension (support), then the weight of the body will not be equal to the force of gravity.

Consider an example. Let a body of mass m lie on the floor of an elevator whose acceleration a is directed vertically upwards (Fig. 107). We will assume that only the gravity force m g and the floor reaction force N act on the body. (The weight of the body does not act on the body, but on the support - the floor of the elevator.) In a reference frame that is stationary relative to the Earth, the body on the floor of the elevator moves together with lift with acceleration a. In accordance with Newton's second law, the product of a body's mass and acceleration is equal to the sum of all forces acting on the body. Therefore: m a = N - m g.

Therefore, N = m a + m g = m (g + a). This means that if the elevator has an acceleration directed vertically upwards, then the modulus of force N of the floor reaction will be greater than the modulus of gravity. Indeed, the floor reaction force must not only compensate for the effect of gravity, but also give the body an acceleration in the positive direction of the X axis.

The force N is the force with which the elevator floor acts on the body. According to Newton's third law, the body acts on the floor with a force P, the modulus of which is equal to the modulus N, but the force P is directed in the opposite direction. This force is the weight of the body in the moving elevator. The modulus of this force is P = N = m (g + a). In this way, in an elevator moving with an upward acceleration relative to the Earth, the modulus of body weight is greater than the modulus of gravity.

Such a phenomenon is called overload.

For example, let the acceleration a of the elevator be directed vertically upwards and its value is equal to g, i.e. a = g. In this case, the modulus of body weight - the force acting on the floor of the elevator - will be equal to P = m (g + a) = m (g + g) = 2m g. That is, the weight of the body in this case will be twice as much as in the elevator, which is at rest relative to the Earth or moves uniformly in a straight line.

For a body on a suspension (or support) moving with an acceleration relative to the Earth, directed vertically upwards, the weight of the body is greater than the force of gravity.

The ratio of the weight of a body in an elevator moving at an accelerated rate relative to the Earth to the weight of the same body in an elevator at rest or moving uniformly in a straight line is called overload factor or, more briefly, overload.

The overload coefficient (overload) is the ratio of the body weight during overload to the force of gravity acting on the body.

In the case considered above, the overload is equal to 2. It is clear that if the acceleration of the elevator was directed upwards and its value was equal to a = 2g, then the overload coefficient would be equal to 3.

Now imagine that a body of mass m lies on the floor of an elevator whose acceleration a relative to the Earth is directed vertically downwards (opposite to the X axis). If the module a of the elevator acceleration is less than the module of the free fall acceleration, then the reaction force of the floor of the elevator will still be directed upwards, in the positive direction of the X axis, and its module will be equal to N = m (g - a). Consequently, the modulus of body weight will be equal to P = N = m (g - a), i.e., it will be less than the modulus of gravity. Thus, the body will press on the floor of the elevator with a force whose modulus is less than the modulus of gravity.

This feeling is familiar to anyone who has ridden a high-speed elevator or swung on a large swing. When moving down from the top point, you feel that your pressure on the support decreases. If the acceleration of the support is positive (the elevator and the swing begin to rise), you are pressed harder against the support.

If the acceleration of the elevator relative to the Earth is directed downward and is equal in absolute value to the free fall acceleration (the elevator falls freely), then the floor reaction force will become zero: N \u003d m (g - a) \u003d m (g - g) \u003d 0. B In this case, the floor of the elevator will no longer put pressure on the body lying on it. Therefore, according to Newton's third law, the body will not put pressure on the floor of the elevator, making a free fall together with the elevator. The weight of the body will become zero. Such a state is called weightlessness.

The state in which the weight of a body is zero is called weightlessness.

Finally, if the acceleration of the elevator towards the Earth becomes greater than the acceleration of free fall, the body will be pressed against the ceiling of the elevator. In this case, the weight of the body will change its direction. The state of weightlessness will disappear. This can be easily verified by pulling down the jar with the object in it sharply, closing the top of the jar with the palm of your hand, as shown in Fig. 108.

Results

The weight of a body is the force with which this body acts on a carrier or support, while being stationary relative to the suspension or support.

The weight of a body in an elevator moving with an upward acceleration relative to the Earth is greater in modulus than the modulus of gravity. Such a phenomenon is called overload.

The overload coefficient (overload) is the ratio of the weight of a body during overload to the force of gravity acting on this body.

If the weight of the body is zero, then this state is called weightlessness.

Questions

1. What force is called the support reaction force? What is body weight?
2. What is the weight of the body?
3. Give examples when the weight of a body: a) is equal to the force of gravity; b) is equal to zero; c) more gravity; d) less gravity.
4. What is called overload?
5. What state is called weightlessness?

Exercises

1. Seventh grader Sergei is standing on the floor scales in the room. The arrow of the device was set opposite the division of 50 kg. Determine the modulus of Sergey's weight. Answer the other three questions about this power.
2. Find the g-force experienced by an astronaut who is in a rocket rising vertically with an acceleration a = 3g.
3. With what force does an astronaut of mass m = 100 kg act on the rocket indicated in exercise 2? What is the name of this force?
4. Find the weight of an astronaut with mass m = 100 kg in a rocket, which: a) stands motionless on the launcher; b) rises with an acceleration a = 4g directed vertically upwards.
5. Determine the moduli of forces acting on a weight of mass m = 2 kg, which hangs motionless on a light thread attached to the ceiling of a room. What are the modules of the elastic force acting from the side of the thread: a) on the weight; b) on the ceiling? What is the weight of the kettlebell? Hint: use Newton's laws to answer the questions.
6. Find the weight of a load of mass m = 5 kg, suspended on a thread from the ceiling of a high-speed elevator, if: a) the elevator rises uniformly; b) the elevator descends evenly; c) the elevator going up with a speed v = 2 m/s started braking with an acceleration a = 2 m/s 2 ; d) descending down with a speed v = 2 m / s, the elevator began braking with an acceleration a = 2 m / s 2; e) the elevator started moving up with an acceleration a = 2 m/s 2; f) the elevator started moving down with an acceleration a = 2 m/s 2 .