A material point moving rectilinearly and uniformly

A material point moves rectilinearly and uniformly with speed V0 along a smooth horizontal surface. At some point, it hits a horizontal rough surface, the friction coefficient of which linearly depends on the distance traveled x: f=ax, where a is a constant coefficient. It is necessary to determine the braking distance of a material point.

Let us consider the movement of a material point on a rough surface. Since it moves uniformly, its speed does not change, i.e. V=const. The friction force acting on a material point is equal to Ffriction=fN, where N is the support reaction force, equal to the weight of the material point: N=mg. Then Ffriction=fmg.

Since the friction coefficient linearly depends on the distance x traveled, then f=ax. Substituting this expression into the formula for the friction force, we obtain: Ffriction=amgx. According to Newton's second law, the force acting on a body is equal to the product of mass and acceleration: F=ma. Consequently, the acceleration of a material point on a rough surface is equal to afriction=gx.

The braking distance can be found using the formula for uniformly slow motion: S=(V^2-V0^2)/(2a). Since the material point moves uniformly before it hits the rough surface, the initial speed V0 is equal to the speed on the smooth surface V0. The final velocity on a rough surface is zero, since the material point will stop. Acceleration a is equal to the braking acceleration afriction: a=afriction=gx. Substituting all the values ​​into the formula, we get: S=V0^2/(2gx).

So, the braking distance of a material point is equal to S=V0^2/(2gx), where g is the acceleration of gravity, x is the distance traveled on a rough surface, and V0 is the initial speed of the material point on a smooth surface.

Task 10731 solved.

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The electronic textbook on physics presents the problem of a material point moving rectilinearly and uniformly with a speed V0 along a smooth horizontal surface, which falls on a horizontal rough surface, the friction coefficient of which linearly depends on the distance traveled x: f=ax, where a is a constant coefficient. It is necessary to determine the braking distance of a material point.

To solve the problem, use the formula for uniformly slow motion: S=(V^2-V0^2)/(2a), where V is the final speed on a rough surface, a is the braking acceleration equal to the friction acceleration on a rough surface, V0 is the initial speed on a smooth surface.

The acceleration of a material point on a rough surface is equal to afriction=gx, where g is the acceleration of free fall, x is the distance traveled along the rough surface. Substituting this value into the formula for the braking distance, we obtain S=V0^2/(2gx).

So, the braking distance of a material point is equal to S=V0^2/(2gx), where g is the acceleration of gravity, x is the distance traveled on a rough surface, and V0 is the initial speed of the material point on a smooth surface.

A detailed solution to the problem includes the formulas and laws used in the solution process, the derivation of the calculation formula and the answer. The electronic textbook on physics is a convenient and accessible way to study physics.

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This product is a material point that moved rectilinearly and uniformly at a speed V0 along a smooth horizontal surface until it hit a horizontal rough surface. The friction coefficient of this surface depends linearly on the distance traveled x and is equal to f=ax, where a is a constant coefficient. It is necessary to determine the braking distance of a material point.

To solve this problem, you can use Newton's laws. According to Newton's second law, the friction force Ftr acting on a material point is equal to the product of the friction coefficient and the normal reaction force N arising from the surface. Thus, Ftr = fN = axN.

According to Newton's third law, the friction force acting on a material point is directed opposite to the direction of its motion. Thus, the friction force will be directed in the direction opposite to the direction of the velocity of the material point.

As a result, using the equation of motion of a material point and the equation for the friction force, one can obtain an equation for the stopping distance of a material point on a rough surface. The solution to this problem can be obtained by integrating the equations of motion and friction force. The answer will be the resulting braking distance of the material point.

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