High School Physical Mechanics, Seek Explanations, High School Physical Mechanics, Seek Detailed Exp

When you use the resultant force, the net force experienced by b is not 20n, because you undercalculate the pull force of a rope against b, and this pull force is used to provide upward acceleration to a by changing the direction of the fixed pulley. It is precisely because of this undercalculated force that your later understanding is incorrect.

It can also be done with the resultant force, but the reasonableness of b is not the 20n you say. Because b is subjected to the downward gravitational force and the pulling force of the rope, the gravitational force is 30n, and the pulling force is not equal to the gravitational force of a, because a has an upward acceleration. Therefore, the tensile force on the rope must be set to t, and the force analysis of AB is respectively

a:t-ga=ma*aa，b：gb-t=mb*ab。

Because the rope is not extensible, aa=ab. Then solve t, and then calculate the net force of b. Or if you understand the natural coordinate system (university general education), you can directly choose the natural coordinate system along the rope, and then list the system of ab Newton's second law.

First of all, your question: because the net force of b in the course of movement is not twenty oxen. Understood this way:

A is accelerated upward by the rope tension, so the rope tension is greater than 30 N, that is, the net force of b is less than 20 N, and it cannot be calculated by your formula: because the mechanical energy of the system is conserved, so mb*g*h=1 2(ma mb)v*v ma*g*h v=self-calculation (2) and then because A is only subject to gravity, the work of gravity is equal to the amount of change in kinetic energy (v has been found).

1, the potential energy of b is not all given to a, because b has a velocity to do work on the ground when b hits the ground.

2. The kinetic energy of the change is 1 2*5v2, which should not be equal.

The first is correct, which is a condition in which the two forces are equal; The second mistake is that the two forces acting on the object will cause deformation on the object, not that it will not have any effect on physics; The third mistake is related to the three elements of force, size, direction and point of action, and the effect may be different for different points of action; The fourth correct, the analysis is the same as the three, the action point is not the same, it may be forward before, and then upward.

1. That's right. Force is a vector quantity, so it is necessary to have both direction and magnitude.

2. That's right. You stand on the ground, subject to gravity and support. It's in equilibrium.

3. Not necessarily. It is different if the object is subjected to other forces.

The effect of a forward force is applied to an object in the horizontal plane, and the effect of the force is not the same with friction on the inclined plane and without friction.

4. Yes. The effect of the force is determined by the point of application and the magnitude of the force.

For example, when you push the door, you push the side of the doorknob, not the side of the hinge.

A and B have always remained stationary, so the acceleration of the two is the same, a = f (ma mb) 1 (the unit is saved for you, remember to add), a motion is subject to a force, that is, b to a friction, for a, a f friction ma, so f friction a ma 1 . The frictional force is in the same direction as the direction of motion of a and b.

Collapse, everyone will, I've been out of school for five years, and I'm afraid I'm wrong.

Judgment method: Look at the x-axis of the coordinate system, because f*sin37>f*cos37 (:f=mgcos37, f=umgcos37

For the resultant force to be zero, the directional afterburner of f*cos37 must be obtained.

After simplification: both proof: sin37>ucos37

Proof that since the object can slide on its own on the inclined plane, the component force of gravity along the inclined plane is greater than the maximum static friction force, and the maximum static friction force is considered to be equal to the sliding friction force).

then mgsin37>umgos37

Hence sin37>ucos37

Since the inclined plane does not move, it cannot be calculated based on the friction factor. Here you still have some concepts to confuse the friction factor: the ratio of friction to pressure when an object is moving, and the key is when it is in motion.

Here the state of motion of the inclined plane does not change, and the resultant force is 0, which is the root of this problem.

Here the external force of the inclined plane has the force and friction force given to him by the object. Therefore, finding the frictional force becomes finding that the force force given to him by the object acts on each other, so the force on the object is equal to that of the object on the inclined plane.

Here I changed the concept again, it became to find the resultant force of the object, analyze it step by step, or figure out the concept first, and then have a model in my head to solve the problem, otherwise physics is not good to learn, I wish you a happy study.

In fact, it is enough to analyze the force of block 1, which is affected by the spring tension force and the friction force given by the ground in the horizontal direction. Since 1 moves at a uniform speed, the net force in its horizontal direction is zero, i.e.:

um1g=xk (where x is the amount of tension of the spring), and x=um1g k is directly solved

It is also known that the original length of the spring is l, then the distance between blocks 1 and 2 is equal to l+x=l+um1g k

The voltage from the voltage source to r2 is one half of the sine wave whose amplitude is half of the sine wave with the amplitude of half of the voltage source's amplitude (the image is the upper or lower part of the x-axis of the sine wave), so with the formula w=tu 2 r, the current on r2 is w=tu 2 r and divided by 2 (because he only has current to do work in half the cycle of the sine wave, although the voltage and current here are not constant values, But you only use this formula to analyze, and it doesn't bring in mathematical calculations, so it's no problem, you can also take the voltage here as an average value), but the work done by the RMS in this period w=tu 2 r The two works are equal to the effective voltage is equal to 18V, so choose A

This one? Looks like a Hengshui golden roll? (Delusion?) Let's talk about it.

The crux of the topic is:The frictional force and the direction of motion are not in the same straight lineAbove.

The first question is, the friction here is the first placeSliding friction。So there is no sudden change (especially in direction), even if there is a tendency to move, it does not care, that is, the friction starts in the same direction as the conveyor belt and is perpendicular to the force. So f directly provides acceleration without subtracting friction.

I said it beforeThe direction of friction is the key to this question。Let's use it firstRelative velocity(relative to the surface of the conveyor belt) to find the direction of friction (importantThe magnitude of friction is the maximum friction, and then the frictionpointsSolutionalongAbsolute movementThe direction is the power of friction.

The second question is to find the relationship between velocity in the f direction and the velocity with time, the vector triangle of velocity and the vector triangle of velocityThe shape is the same(I used the Pythagorean theorem for this place, which is simple and crude.) We're going to find it based on the relationship of speedThe magnitude of the frictional force in the f direction。To get this f to represent the variation of the frictional component obtained, we only need to follow Newton's second law.

Solve with f-f=am to get the equation f about t.

What do they have in common, no matter where they fall?

That is, the motion is decomposed in the direction of the vertical bevel along the inclined plane (as opposed to the horizontal and vertical direction in the usual analysis).

And when decomposed, their displacements in the direction of the vertical slope are equal.

In the direction of the perpendicular inclined plane, there is a component of gravitational acceleration (fixed) and a component of initial velocity. Obviously, this component is the largest when the initial velocity is perpendicular to the bevel. It takes the shortest time.