High school physics is easy to make mistakes, give high school physics easy to make mistakes, classi

It's up to you to accumulate it yourself.

Someone else's may not necessarily be yours.

Every time you make a mistake, write it in the wrong question book.

After a while, it's a good error-prone question book.

A collection of typical examples of high school physics (1).

Mechanics section. 1. As shown in Figure 1-1, the two ends of the 5-meter-long rope are tied to the top ends of two poles A and B erected on the ground at a distance of 4 meters. Hang a smooth, lightweight hook from the rope.

It hooks an object weighing 12 Ox. In equilibrium, the tension in the rope t=

Analysis and Solution: This topic is the problem of the balance of forces. The basic idea is:

Selecting objects, analyzing forces, drawing force diagrams, and column equations. For equilibrium problems, different methods can often be used according to the conditions given by the topic, such as orthogonal decomposition method, similar triangles, etc. Therefore, there are multiple solutions to this problem.

Let the angle between the string and the horizontal be , and it can be seen from the equilibrium condition: 2tsin = f, where f = 12 N.

Extending the rope, from the geometric conditions in the figure: sin = 3 5, then substituting the above equation can obtain t = 10 N.

Solution 2: The hook is subjected to three forces, and it can be seen from the equilibrium condition that the resultant force f' of the two tensile forces (equal in magnitude t) is in the opposite direction when f is equal in magnitude.

Cattle. Think about it: if the right end rope A is moved down the rod a bit, does the tension on the string change?

Example 1: A car with 10

Sudden braking after 5 minutes of exercise at the speed of ms. For example, the braking process is to do a uniform speed movement, and the size of the liquid degree is 5m s

What is the distance the car travels in 3 seconds after braking?

Wrong solution] Because the braking process of the car does a uniform deceleration linear motion, the first speed of the object is accelerated vm s.

Misinterpretation Reason] There are two reasons for the above error. One is that the physical process of braking is not clear. When the velocity decreases to zero, there is no relative motion between the car and the ground, and the sliding friction becomes zero.

Second, the physical meaning of the displacement formula is not well understood. The displacement s corresponds to time t, during which a must exist, and when a does not exist, the calculated displacement is meaningless. because of the ignorance of the first point is to think that athere is forever; Because of the incomprehension of the second point, there is a contemplation of when a does not exist.

Analyze and answer] Draw a motion sketch 1-1 according to the topic. Let the warp time t velocity be reduced to zero. According to the formula v for the speed of linear motion of uniform deceleration

Vat has 0=10-5t solution to t=2s because the car is in 2s.

Comment] Physics problems are not simple calculation problems, when the results are obtained, we should think about whether it is related to.

s=-30m, which is not consistent with reality. It is necessary to consider whether there is a problem that is inconsistent with reality in the application of the law.

This problem can also be solved using images. The braking process of the car is a uniform deceleration and linear motion. From this VA we can see the triangle V

The area enclosed by the OT is the displacement within 3 seconds of the brake.

I've never seen this question before ( o Ah!

Select b to fall along the curve of the flat throw. Just drop the top to the far left side without touching the edge of the table.

The fall time t= (2h g) and the distance is r

vt=r yields v=r g 2h

By the formula: a=v2r

Obtainable Cover Dust Journey: r=v 2 a=150*150 (6g)=3750 g (g is the gravity plus the speed of the cover).

1min=60s, 1s=30 times, 1min=1800 times. Because there is no feeling of rotation, the blade may rotate 120N(N Z) at a time, when N = 1, 1800 120 = 216000°, 216000° 360° = 600R, so a to the cavity early limb, other options are counted by themselves.

(1) Set the horizontal to the right as the positive direction.

Conservation of momentum: mv0-mvx=0, vx=mv0 m

2) By the kinetic energy theorem:

mgs=1/2mv0^2+1/2mvx^2,s=1/2v0^2(1+m/m)/μg

s is the sum of the displacement of the small wooden block and the wooden block, you may wish to draw a drawing yourself, mgs is the total work done by the frictional force on the small wooden block and the wooden block).

Since the small wooden block of mass m is placed on a long board of mass m**, in order to prevent the small wooden block from falling from the wooden block, there should be S 1 2L

So there is: l 2s = 2 [1 2v0 2(1+m m)] g i.e. l v0 2(1+m m) g

1.Conservation of momentum: vx=mv0 m

2.Relative motion: l=m(m+m) m2g

As shown in the figure, because it is a smooth horizontal plane, so the two wooden blocks are only subjected to one frictional force, and the two wooden blocks stop at the same time to indicate that the time t is equal, for two wooden blocks, let their acceleration be a1, a2, then a1 = g, the friction force of the large wooden block is f= mg given by the small wooden block, the acceleration of the large wooden block is equal to a2= mg m, the two times are equal, then vx a1 = v0 a1, and the substitution calculation obtains vx=mv0 m

If both blocks stop moving at the end, the small blocks do not fall. Let the sliding distance of the wooden board be s2 and the sliding distance of the small wooden block is s1, then s1+l 2=s2. There is no solution by the calculated formula.

If two wooden blocks reach the same velocity at the same time, both move together. Count the rest yourself.

In the force analysis, only the horizontal direction is considered, and the three objects A, B, and C are only subjected to centripetal force and friction. The condition for an object to be at rest is that the centripetal force is equal to the static frictional force. The centripetal force increases with the increase of rotational speed, and when the centripetal force is greater than the maximum static friction force, the object produces centrifugal motion.

According to the title, if the coefficient of dynamic friction is the same, then the maximum static friction (set to f) is proportional to the mass. So: fa = 2fb = 2fc

According to the law of circular motion, the formula for centripetal force and centripetal angular velocity is:

f = mv r = m r (m is the mass, r is the radius).

a = v²/r = ω²r

1) All points on the table have the same angular velocity, then the centripetal angular velocity is only related to the radius, because 2ra = 2rb = rc, so option a is correct;

2) When at rest, the frictional force is equal to the centripetal force. fa = (2m)ω²r；fb = mω²r；fc = m (2r), so option b is correct.

3) The sliding condition is that the centripetal force is greater than the maximum static friction force. From Fa = 2FB and Fa = 2Fb, it can be seen that the maximum static friction force and centripetal force of A are both twice that of B, so A and B will slide at the same time. Option C is incorrect.

4) From fb = fc and fb = fc, it can be seen that b and c have the same maximum static friction force, but the centripetal force of c is 2 times that of b, so c slides first. Option d is correct.

d, examines the use of thinking methods and formulas that are divided into parts and viewed as a whole.

For f=mrw f is proportional to r if w is constant, and f=m (v r) f is inversely proportional to r if v is constant.

Here we see why the two objects are still balanced at first? Apparently the tipping point has just been reached, and from the first of the two formulas we think that there are ropes in the whole now, but they are subjected to different forces. And the angular velocity is the same, so we use the first formula f=mrw

Balance is achieved from the whole question. The force experienced is friction. The maximum static friction is achieved from the question.

Re-analysis of the monomer angular velocity is unified, found that the outside of a single A to maintain the balance of the force is large, a single surface B to keep the balance of the force is small, but the actual friction is the same, and the single in this resultant force is formed by the rope and static friction, so we also know the direction of the pulling force, to A is to pull to the center of the circle, B is to pull outward.

B was originally immobile, but was pulled by A. Therefore, after breaking, the friction at B will become the size of the previous resultant force, which is actually smaller (not from static friction to sliding friction).

And A should not slide, unless the resultant force received by A is greater than the frictional force it has been subjected to, it can slide constantly. Obviously, it is already the maximum static friction, and there is no pulling force, and the resultant force is its own friction, so it slides.

Let the force of A on B be fSo b to a is also f

The centripetal force of a = mag + f = mraw

The centripetal force of b = mbg-f = mrbw

After the break. The maximum centripetal force of a and the maximum centripetal force of b are relatively stationary.

d. I can't see the ...... of the picture clearly

The one on the outside slips and the other still moves in a circular motion with the disc.

If I'm not mistaken, it's D.

Because the radius of the circular motion of object A is larger, a greater centripetal force is required, and the rope can no longer provide tension after it is burned, and the frictional force given to A by the disc alone is not enough to provide centripetal force, so sliding occurs.