Senior 1 physics questions about conservation of mechanical energy

Because the object moves in a straight line at a uniform speed from the stationary to the time when the object moves with the conveyor belt, then s=(v 2)*t and the conveyor belt velocity is constant, then his distance or displacement is s=vt, which is twice that of the object, and the first two should be the same as mv 2 2. The last one should be MV 2

Normally we think of the work done on an object as equal to the force multiplied by the displacement to the ground.

The displacement here refers to the displacement of the counterpart.

Answer: A Because the mechanical energy should be conserved, there is only gravitational potential energy at the highest point, so it is the same as the original!!

Upstairs is all the right solution.

During the relaxation phase of the rope, the mechanical energy of the ball is conserved because the rope has no force and the ball is only free falling by gravity. But at a certain time, when the position of the ball and point A are exactly equal to the length of the rope, (at this time, the position of the ball and the line of a is definitely not in the vertical direction, but at an angle with it), the rope is straight, because the ball velocity is vertical downward, and the rope can not be extended, so this speed is decomposed into a tangent velocity perpendicular to the direction of the rope and a normal velocity along the rope extension line, due to the existence of rope tension, In a very short period of time, the impulse generated by the rope tension will cancel out the momentum of the normal ball, and then only the tangential velocity is left (this cancellation process can be considered instantaneous under normal circumstances), so in fact, the kinetic energy of the normal direction is gone, so the mechanical energy is naturally not conserved, because the rope tension does negative work, and the mechanical energy of the ball decreases. After that, the ball moves in a circle around A, and if it is very small (within 5° in middle school, it seems), then we can approximate the circular motion with a single pendulum motion.

In the extreme example, the initial position of the ball is directly below point A, and the rest of the conditions remain the same, you can see that in this case the mechanical energy of the ball is not conserved.

Your teacher said very professionally, first of all, the ball falls freely, when the rope is tightened, it happens that the angle between the rope and the horizontal direction is also 30 degrees, at this time, the speed direction of the ball is vertical downward, at the moment when the rope is tightened, because the rope can not be extended, so the speed component along the direction of the rope is offset by the work done by the tension of the rope, it can also be said that the impulse of the rope to the ball cancels out this speed component, just because the action time is too short, it is not detected. This will be learned in the momentum theorem of the second year of high school, after which the ball only has a velocity component perpendicular to the rope, its motion process is similar to a single pendulum, the rope has not done work, knowing that the ball reaches the lowest point, the mechanical energy of the ball in this path is conserved, and the whole process of work can be strictly calculated. You can do the math.

It's not a problem. It's just that I think the teacher's example questions here are very distinctive.,It's a very experienced physics teacher.,Hate to admire,,This example tells us that don't rush to start in the analysis of physics problems.,We must analyze it carefully.。。。

It is not conserved, the free fall is downward, and then when the rope is straightened and held, the direction of velocity is changed to the tangent direction of the circle. You decompose the velocity at that time into the direction along the tangent and the perpendicular tangent, then the speed in the perpendicular tangent direction becomes 0 because the rope is straightened (this is why the mechanical energy is not conserved, the speed in that direction is absorbed by the rope, in fact, it is not absorbed instantly, it is converted into the vibration of the rope, and then slowly neutralized by the resistance, but the problem is needed, it is regarded as suddenly gone). Then the velocity along the tangential direction can be understood as the conservation of mechanical energy.

Therefore, the whole is not conserved, because the velocity of the vertical tangent is gone. But that point is conserved before and after. Before it was free fall, conserved; This is followed by oscillation, which is also conserved.

If you don't understand, you can ask.

v2 According to the kinetic energy theorem, let the maximum height that the object can reach be h Ascending process: 0-1 2mv1 square = -mgh-fs (1) Since a is the midpoint of the object reaching the maximum displacement, let the object velocity at this time v1 2mv squared -1 2mv1 square = -mgh 2-1 2fs (2) According to 1,2, 1 2mv square = 1 4mv1 square = 1 2mgh + 1 2fs

and because f,s are both greater than 0So at point a, the kinetic energy is greater than the gravitational potential energy (1 2mgh), while the kinetic energy gradually decreases, and the gravitational potential energy increases, so the point where the kinetic energy is equal to the gravitational potential energy is located above point a.

The title is wrong! v2 can't be greater than v1!

Conservation of mechanical energy.

The angular velocity of a b is the same, and the radius of motion a is 1 2v= r of b.

Vertical: vb=2va

mgl+mgl/2=mvb²/2+mva²/2vb=2√(3gl/5)

1.Let the maximum height that the ball can rise from after the first collision with the ground is h, then the kinetic energy theorem is obtained: mg(h-h)-kmg(h+h)=0

So the solution is h=(1-k)h 1+k

2.The total distance traveled by the ball from the beginning of release to the end of bouncing is s, and the whole process is obtained by the kinetic energy theorem mgh-kmgs=0

So get s=h kok!

The conversion of all the potential energy of the weight into kinetic energy is the maximum, and you just have to take into account the fact that he did the most work at the above time.

is the square of mg1 2h-mg1 4h=1 2*(4m)v mg1 2h=1 2*8mv 2, then v = root number (gh 8).

If you want to go deeper into the point of gravity and convert the most to that point, you can also do the same calculation, after pumping the lid, the right height is x on the right, and the left drops to the height y, then x+y=h is conserved according to kinetic energy:

1 2*(4m)*v squared = mg1 2h-x h*(mg)*1 2x-y h*(mg)*1 2y Finally, v=root number 2xyg 2h to make v to a maximum of 2xy<=x squared + y squared, so when x=y is 2xy maximum, that is, v maximum, i.e., x=y=h 2, the result is also the root number (gh 8).

This must be given points, guy,

When the two liquid levels are at the same height, the gravitational potential energy is converted into kinetic energy to the greatest extent, so the velocity is the greatest at this time.

The liquid level on both sides is equivalent to the height of the right side 1 2h The center of gravity of the liquid column decreases, and the mass of this part of the liquid is m, then the total mass of the liquid is 8m, according to the law of conservation of mechanical energy:

mg1 2h = 1 2 * 8mv 2, then v = root number (gh 8).

You see, when the liquid level is equal, the pressure on both sides is equal, the pressure is equal, and the cover plate is removed, there is no gas pressure, then the resultant force is not zero, the resultant force is zero, and the acceleration is zero, then the velocity is the maximum!

The object is the most stable when the energy is the lowest, the energy is not the lowest when the object is in the unstable state, and the most stable potential energy is the smallest and the kinetic energy is the largest when the liquid level is at the same height.