A high school physics problem, kinetic energy theorem and conservation of mechanical energy, ask the

Because the title doesn't say that b and d are equal to each other, I don't know if the same height can be used as a condition.

Solution: (1) After the ball is thrown, the ball enters the track along the tangent direction from the C end of the track, indicating that the end velocity of the ball should be along the tangent direction of the C point, and the speed of the flat throw is decomposed, and the geometric relationship obtains:

The vertical component of point C velocity: vy = v0tan45° = 6m s

The sub-motion in the vertical direction is the free fall motion, t=vy g = horizontal direction to do the flat throwing motion, l=v0t=

2) From the law of conservation of mechanical energy, there is 1 2 mvC2 + mg(r-rcos45°) = 1 2 mva2 according to the centripetal force formula: fa-mg=mva2 r

Solution: fa=

3) Let the ball reach point b, according to the law of conservation of mechanical energy, there is 1 2 mV02 + mg (h-r-rcos45°) = 1 2 mvb2 solution to obtain: vb = 38 m s gr

So you can get to point B.

Answer: (1) The horizontal distance l between the point d of the ball throwing point and the c end of the circular orbit is;

2) When the ball passes through the lowest point of the orbit a, the force of the orbit is fa;

3) The ball can reach the highest point b

According to the angle between the vertical and horizontal directions of the velocity is 45°, the vertical velocity and the falling time t= can be calculated

The fastest to say it clearly....I didn't say that I was equal to the height....

Idea: Kinetic energy theorem.

It is said that "the work done by the combined external force is equal to the change of kinetic energy", the combined external force includes the conservative external force and the non-conservative external force, in which the work done by the conservative external force is equal to the change of the negative potential energy, and the change of the mechanical energy is moved to the side of the kinetic energy, and the left becomes the non-conservative external force to do the work, if this term is equal to zero, the mechanical energy remains unchanged, which is.

The law of conservation of mechanical energy.

Can be pushed out.

However, this is a special case of conservation of mechanical energy.

What if I have a spring?

Your kinetic energy theorem does not necessarily lead to the conservation of mechanical energy.

Hehe. As for the derivation, I personally think that when I wrote it, it seemed to be written about the kinetic energy theorem, but in fact, it was the conservation of mechanical energy.

If you don't bother, I'll look at what the brother said below, it's good

1 The change in the total gravitational potential energy is only related to the work done by gravity, and gravity does the work of -mgh, so the gravitational potential energy increases mgh, a is wrong.

The change in physical kinetic energy is only related to the work done by the resultant external force, which is ma, i.e., mg2, in the direction downward, so the work done by the resultant external force is -mgh 2, and the kinetic energy decreases mgh 2, b correctly.

The mechanical energy of an object is the sum of kinetic energy and potential energy, the gravitational potential energy increases mgh and the kinetic energy decreases mgh2, then the mechanical energy increases mgh2, c is true, d is false.

The mechanical energy involved in this question is gravitational potential energy (ep) and kinetic energy (ek) and has ep=mgh

ek=mv'2-mV2=-mH2 (by v'-v = 2as to get).

e=mgh-mgh/2=mgh/2

Calculate the kinetic energy at the bottom: analyze the force on the iron block, the upward orbital elastic force, the downward gravity mg, the resultant force is the centripetal force, and the kinetic energy =

Then the energy lost is known by the conservation of mechanical energy.

22.Taking the small ball as the research object, it is affected by gravity and tension, and the pulling force does not do work.

1. The expression for the conservation of mechanical energy mgl(1-cos60°)=1 2mv back 22, the kinetic energy is fixed.

Answer expressions. mgl(1-cos60°)=1 2mv 2-03, energy conservation expression. mgl(1-cos60°）=1/2mv^2v=[2gl(1-cos60°)]1/2=2m/s

First, analyze the forces of a and b.

A is subjected to gravity and the pulling force of the rope, and the resultant force is downward, you can mentally calculate its acceleration a=10 3m s2, the force of b is the force of gravity and the rope tension, the resultant force is upward, calculate its acceleration a=5m s2,1, according to the formula of acceleration, vt 2-vo 2=2as, this is the formula in the book, here s=1, vo = 0, vt can be calculated, at this time vt = 2 root number 15 3, this is the instant speed of a landing.

2. Analyze b, b wants to reach the maximum velocity, that is, when the acceleration is 0, that is, when a lands, then according to the formula vt 2-vo 2=2as, s is the same as the previous one, because it is a pulley, vo = 0, v maximum = root number 10

3, this question is a bit strange to ask, B doesn't seem to have landed, B's movement process is to accelerate upwards first, and then by the action of gravity, decelerate, but still upward, and finally become 0, then accelerate downwards If the energy consumption is not considered here, until B returns to the ground at a speed of 0, the same as the beginning of the state, and then repeat this process. You can talk to me about it.

The recommended answer is wrong! (From the first overall acceleration is wrong, a=5 3 m s 2, all the following are wrong).

Solution: (1) AB system: A falls in the H process, the mechanical energy is conserved, and the ground is taken as the zero potential energy surface MGH=MGH+

The simultaneous solution is v=2 m s

2) B: The process of throwing vertically to the highest point, the same goes for it.

mgh+ is the height of the highest point from the ground).

Substituting the solution gives h1= m

3) B: The process from the highest point to the ground, the same is true.

mgh1 = solution v1 = 2 6 m s

Explanation: It is to solve the problem with the conservation of mechanical energy according to your requirements, in fact, (2) and (3) are simpler to solve with the laws of kinematics.

A false, the rope must have a component in the horizontal direction;

B Yes, if you want object B to be stationary horizontally, there must be a force to balance the horizontal component of the rope, and this force can only be frictional and equal to the horizontal component.

c False, there must be friction in the above analysis, and the condition for friction is deformation and extrusion, so there must be support on the ground.

d Yes, you can find out by analyzing it yourself.

1. Regardless of air resistance, the energy of the system is conserved in the whole process, and the work done by the springboard on the athlete is the kinetic energy of the athlete when he enters the water, that is, ek=1 2*50kg*(10m s) 2=2500j

2. At the highest point, the mechanical energy of the system is only the gravitational potential energy of the athlete, that is, EP=EK=2500J, and EP=MGH, then H=2500J 50KG 10M S2=5MHowever, this 5m is the distance from the water, and since the springboard is 1m from the water, the athlete is 4m from the springboard.

1: The gravitational potential energy and the work done by the springboard on the athlete are converted into kinetic energy, and the horizontal plane is selected as the zero potential surface.

E plate + mgh = 1 2 * mv 2

E plate = 2000J

2: The work done by the springboard on the athlete is converted into gravitational potential energy.

E plate = mgh

h=4m

As shown in Figure 11, a small ball with mass m= is moved at the initial velocity v from point o at the right end of the horizontal platform. Thrown horizontally, the ball flies away from the platform and falls into the vertical smooth circular orbit ABC along the tangent line from point A, and passes through the highest point C along the track, the shape of the circular orbit ABC is the circle of radius R= M and the arc of L270 in the upper left corner is truncated, and CB is its vertical diameter, (sin530= cos530=, gravitational acceleration g takes 10m s2) find:

1) The velocity of the ball through point C;

2) The pressure of the ball on the track when the ball moves to the lowest point of the track b;

3) The vertical height h from point O to point A at the end of the platform.

(1) Because the ball happens to pass through point C. So gravity provides centripetal force. f=mg=mv2 r, so v=root, gr=5m s

1) The ball passes exactly through the highest point c along the orbit, and gravity provides the centripetal force, i.e., mg=m, and the solution is vc==5m s.

2) From point B to point C, the ball is analyzed by the law of conservation of mechanical energy MVC2+2Mgr= MVB2, and the force analysis of the ball is carried out at point B, and Newton's second law has fn — mg=m and the solution is fn=.

According to Newton's third law, the pressure of the ball on the orbit is.

The ball with a mass of m is fixed at the upper end of a smooth light and thin rod, and the thin rod is kept upright by a smooth limit hole. A concave groove with a mass of m 2m is placed on a smooth horizontal plane, the smooth inner surface of the groove is shown in the figure, the ab part is an inclined plane, and the horizontal plane is 30°, the bcd part is the arc surface of radius r, and the two sides of ab and bcd are tangent at b. Let the lower end of the thin rod come into contact with the left edge of the groove at point A.

Now release the ball and ask for:

1) When the lower end of the light thin rod slides to the lowest point c of the groove, what is the speed of the groove;

3) When the lower end of the light rod slides to point B, the velocity of the ball and the groove is great.