High school physics questions for guidance, high school physics questions for teaching

To make ab slide relatively, there is sliding friction between ab, and the magnitude of the sliding friction between ab is gravity multiplied by a = 1nThe sliding friction between b and the ground is at least 1n+1n+6n=8n, and the second question f is at least 4n+4n+3n=11n, sorry, I don't know how to type mathematical expressions. 1n+1n+6n=8n means the friction between ab, the tension of the rope and the friction between b and the ground.

Let's look at this problem, let's start from simple to complex, obviously analyze A first and then analyze B, for A, by 4 forces, gravity, support force, friction, tension, want to move need to be balanced by force so that we can get the pull of the rope = B to the sliding friction of A. F-rope = f=1n

Then we analyze the forces on B, B are pressure, gravity, support, friction of A against B, friction on the ground against B, and tension on the rope.

Let's analyze the magnitude of the force one by one, support force = pressure + gravity = 30n

So the frictional force of the ground facing b = support force * u = 6n

Friction of A to B = Friction of B to A = 1N

Because the pull of the pulley rope is equal everywhere (in high school), the pull force to A = the pull force to B = 1N

So in the horizontal direction the pull force = 6 + 1 + 1 = 8n

In the first question, the ship is always moving in the horizontal plane, there is no displacement in the vertical direction, and the velocity is always 0 without change, so the acceleration is also 0, so there is no need to consider it;

In fact, the second question is quite difficult and quite confusing. Putting this question aside, let's first discuss a simple question: when you take the elevator up 10 floors at a constant speed, you are supported by gravity and the elevator floor, and the resultant force is 0, although your position is raised, your gravitational potential energy also increases, but the resultant force is still 0.

So when discussing the work done by the combined force, don't consider the increase in gravitational potential energy. Coming back to this question, this should be a B, not an A.

In fact, the second question is quite difficult and quite confusing. Putting this question aside, let's first discuss a simple question: when you take the elevator up 10 floors at a constant speed, you are supported by gravity and the elevator floor, and the resultant force is 0, although your position is raised, your gravitational potential energy also increases, but the resultant force is still 0.

So when discussing the work done by the combined force, don't consider the increase in gravitational potential energy. Coming back to this question, this should be a B, not an A.

In the third question, if the energy consumption is not counted, it is impossible for the spring to balance again (stationary), because before reaching the equilibrium position, the gravitational force of the object is always greater than the elastic force of the spring, so it has been accelerating downward, and when the velocity reaches the maximum value when it reaches the equilibrium position, it has kinetic energy, and it will continue to move downward, so the elastic potential energy increased by the spring at this time is not equal to the gravitational potential energy of the object decreased. So you still have to use the formula of elastic potential energy to calculate honestly.

Question 2: Because it is on a smooth inclined plane and there is no friction, the work done by the sedan rush is calculated by the kinetic energy theorem formula as 1 2mv 2-0, that is, the work done by the pulling force. Mess up.

1. No matter how many times the bird flies, the time it flies is always the time of the distance s of car A and B, that is, t=s (v1+v2), and the speed of the bird's flight remains the same, then, the distance of the bird's flight s'=v0*t is s'=v0*s/(v1+v2)

2. No matter how big the string is mentioned above, the most important thing is that the plane suddenly stops, which is regarded as a speed of 0, and the pilot's speed is approximately 0, and the pilot is accelerating to 20m s, the formula v1=v0+at, that is, 20m s=a*, a=200m s 2

1.The velocity of the bird relative to A is vo, then the velocity of the bird relative to the ground is v1 + the time required for the encounter of vo A and B t=s (v1+v2).

Then the distance traveled by the bird is s(v1+vo) (v1+v2), and at the moment of ejection, it can be considered that the vertical velocity of the aircraft is 0, that is, the initial velocity of the person is 0

Analysis: Let a drop of water fall every time t, then 1 2*g*t = 1 can be obtained from the question, and the solution is t= 5 5 s, and the fifth drop of water falls when the first drop of water lands, it can be seen that the time required for the water droplet to land is t=4t, that is, 4 5 5 s, so there is h=1 2*g*t =16m, that is, the eaves height sought is 16 meters.

If you don't understand, you can ask again

The time interval between each two drops of water is equal, set to t, and since it is a free-fall motion, the initial velocity is zero, and the ratio of the displacements passed in a continuous equal time is 1:3:5:

7……There is a time of 4t between the first and fifth drops, so the displacement in these four time periods is 1m3m5m7m, respectively, and the total height is 16m

The water droplets move uniformly with an initial velocity of 0 and an acceleration of g, and because water droplets fall at equal intervals, the displacement ratio of adjacent equal time intervals is 1:3:5:2n-1).

There are four time intervals, and the ratio is 1:3:5:7

So the eaves height is 1*16m=16m

o'1 Signs for judging whether two objects are separated: two objects separate when their acceleration a is not equal.

2.When m and m compress the spring and then ** (not yet to the equilibrium position o), the acceleration of the two is the same (otherwise it will be separated), when passing the o point, m is subjected to gravity, the pressure exerted by m and the elastic force of the spring, then m and m have the same acceleration, when o is reached', the elastic force of the spring is 0, that is, m and m are weightless, and there is no force between the two (about to separate), when the o is to be passed'At the point, m will be subjected to gravity and downward pull so that the acceleration a>g, i.e., the two are separated, so o'points when separating points.

I don't know if it's satisfied.

The maximum acceleration that M can achieve during the ascent (direction downwards) can exceed that of G (in O'above the point), while the maximum acceleration that m can obtain can only reach g (reached after detachment from m), obviously beyond o'After the point, the acceleration of m will exceed g, which means that after that the velocity of m decreases faster than m, so m cannot keep up with m, and the two separate.

o', separation, no force, m falls from a height,。。 Then it moves upwards again

The separation of m and m means that the support force of m to m is zero, that is, the acceleration of m is g, and the acceleration of the two before separation is the same, that is, the acceleration of m at the moment before separation is also g, and the acceleration of the two objects at o is zero, so it is definitely not separated at o but at o'separation.

The upper one, because of the downward elasticity of the original long position, the two are separated.

In the original length position, the downward acceleration is negative, and then the upward acceleration is positive, and after reaching the original length position, the spring force is reversed, and the m acceleration changes from positive to negative.

t=2s/v1

The time of the steamer round trip is accompanied by spikes:

t's (v1+v2)+s (v1-v2)s[1 bopei(v1+v2)+1 (v1-v2)](v1*t 2)[1 (v1+v2)+1 (v1-v2)](v1 2)t guess the silver stupid[(v1+v2)(v1-v2)]v2s t

2/[1/(v1+v2)+1/(v1-v2)](v1+v2)(v1-v2)/v1