High school physics questions, help analyze them

Answer A: Analysis: Because the object can slide at a uniform speed, it means that the resultant force of the supporting force and the sliding friction force on the object is equal to the gravity and the opposite direction, so the resultant force of these two forces (that is, the force of the oblique object) is vertically upward, and the force (resultant force) of the object on the inclined body is vertically downward, so there is no friction on the ground.

When a thrust force is applied to an object, it can be seen through the analysis of the force on the object that the object will move downward with uniform acceleration. However, since the frictional force is proportional to the supporting force, the resultant force of these two forces is still vertically upward, so the force of the object on the inclined body is still vertically downward, so the frictional force of the inclined body on the ground is still zero.

The general feature of this problem: as long as the object is moving at a uniform speed on the inclined plane, no matter how much the force is increased and in what direction, as long as the object still slides along the inclined plane, the friction force of the inclined body on the ground is zero.

I'm sorry, it's hard to draw, so if you draw the friction and support force yourself, you will be able to see it easily. Hope it helps.

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For wooden blocks, the horizontal direction is subjected to a rightward tensile force f and a leftward frictional force f = m2*g. then f-f=m2*a2

For planks, if the horizontal direction is subjected to a frictional force to the right f= m2g, then f=m1a1

For acceleration, a1=a2 is possible, in which case f=(m1+m2)a. In B, A1>A2, then V1>V2, then the wooden block should be subjected to the right friction, the wooden board should be subjected to the left friction, and the wooden board will be evenly decelerated. Here there is a contradiction.

So for acceleration, a is possible, and b is impossible.

For speed, v2>v1, it is reasonable for the block to experience a frictional force to the left and the plank to the right friction to accelerate.

If v1>v2, the block should be subjected to the right friction, and the board should be subjected to the left friction, and the board will be evenly decelerated. Here there is a contradiction. So for velocity, c is possible, d is not.

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Force Analysis:

For the wooden block, the horizontal direction is subjected to the right tensile force f, and the wooden board contact has a leftward friction force f2, the size is not determined, but the maximum is m2g, where is the friction coefficient (assuming that the dynamic and static friction coefficients are the same, are .

For the plank, the horizontal direction is subject to friction at the contact with the wooden block, the direction is to the right, and the size is equal to F2. The contact with the smooth horizontal plane is not clear, whether it is considered that the friction coefficient is 0?. Then there is no friction at the point of contact.

Problem analysis: 1. F is relatively small, less than the maximum friction m2g at the contact between the wooden block and the wooden board, and there is no friction at the contact between the wooden board and the smooth horizontal plane, then the wooden board and the wooden block move to the right together at the same speed and the same constant acceleration, and its value is f (m1+m2). This is shown in Figure A.

2. F is greater than the maximum friction force m2g at the contact between the wooden block and the wooden board, then the wooden block moves to the right with the acceleration of m2g (m1), and the wooden block moves to the right with the acceleration of the value (f- m2g) (m2) relative to the wooden board (note that it is relative to the wooden board), that is, the acceleration of the wooden block is greater than the acceleration of the wooden board. Then it is not possible in the acceleration diagram a and b, only in figure c: it is possible that the velocity of the wooden block is greater than the velocity of the wooden board.

3. F= M2G, critical state, unstable motion (it may be the first article above, which is the state of Figure A, or the second article above, which is the state of Figure C). However, it is not possible to see the states of Figures b and d.

Therefore, the ones that may be consistent with the exercise are (a, c).

In this kind of problem, we should pay attention to the relationship between f and f, when f "maximum friction m2g, f = f, not m2g.

Free fall from a standstill, 4 meters from the starting point in 1 second: i.e. h=1 2at 2 8=a

h4=1/2at2=1/2*8*25=100

If you still don't understand, you can ask me!

Use x=gt to find the t time interval.

x=bc-ab=20cm

t= = choose c