Ask for a physics problem about the kinetic energy theorem

This question examines the kinetic energy theorem, where the work done by an external force is equal to the amount of change in kinetic energy. The work done by the elastic force of the object plus the work done by gravity is equal to the final kinetic energy of the object minus the initial kinetic energy of the object, that is, W bomb + W weight = EK end - EK beginning, W weight = - mgh (gravity does negative work), EK end = 0 (the end velocity is zero, that is, the velocity at the highest point is zero), EK initial = 1 2MV2, and W bomb = MGH - 1 2MV2. In this process, the kinetic energy is converted into gravitational potential energy and elastic potential energy, and the final gravitational potential energy is smaller than the initial kinetic energy, so the elastic force is negative work.

Did you write the change of kinetic energy wrongly, the initial kinetic energy is not zero, and the final kinetic energy is zero, so the kinetic energy variable is negative.

That's right: making the ball reach a height h to form gravitational potential energy has two parts of work, regardless of positive or negative. One is the kinetic energy of the ball 1 2mv, and the other is the work done by the spring to the ball: w, so there is:

1/2mv²+w=mgh

So: w=mgh-1 2mv

Isn't this obviously a choice of A, obviously W is a negative work.

The landlord you listed the formula is very correct, that is, -w-mgh=0-1 2mv 2

But the problem here w is positive, and you ask for the negative work done by the spring, so the answer is, of course, what is -w, not w.

The kinetic energy of the ball is converted into two parts of energy: the potential energy of the ball and the work done on the spring. The work done by the ball on the spring = 1 2mv -mgh.

The work of the spring = — the work done by the ball on the spring, so, the work of the spring = —(1 2mv -mgh) = mgh-1 2mv.

It is negative work, and the kinetic energy changes into elastic potential energy plus gravitational potential energy, so isn't the work done by the spring equal to the elastic potential energy equal to the kinetic energy minus the gravitational potential energy.

It is to consider the negative work of the spring to choose A.

1/2mv^2=mgh-w

Landlord, you all know that elasticity has done negative work, that is, mgh and 1 2mv who is bigger and who is small.

The kinetic energy of the object at point A 1 2mv is its total mechanical energy, in the process of A to C, the kinetic energy is converted into the gravitational potential energy mgh of the object and the elastic potential energy of the spring, so it should be 1 2mv mgh

So, you should choose A

I think it should be explained like this:

According to the function --- the transformation definition.

w=e2-e1

mgh-1/2mv2

So choose A

You miscalculated, the teacher is right.

The electric field force experienced by the small block is greater than the sliding friction force, so in the end the small wooden block will stop at the left end, so it is equivalent to the electric field force does not do work, and all kinetic energy wants to be converted into internal energy, so there is a final answer of 1m

The equation is: fs = 1 2mv squared.

1: You are talking about the work done by external forces, yes, the work done by the combined force only needs to consider: the beginning and end state;

2: But, here we are talking about a pulling force: not an external force; Tension = combined external force + gravity * cos60 (Note: combined external force = pull force - gravity * cos60; )

Should it be: gravity * cos60 or : gravity * sin60, you confirm again,)

The force causes the object to rise the gravitational potential energy at h height and the kinetic energy with velocity v.

It's just the beginning and end states, and you forget that it's increased gravitational potential energy.

You are right, in the kinetic energy theorem, only the kinetic energy of the initial and final states of the object needs to be considered, but the calculation formula derived from the kinetic energy theorem in this problem should be w-mgh=1 2mv2, w is the work done by the tensile force, and -mgh is the negative work done by gravity, so the result of the answer is obtained after the equation is shifted.

Quite simply, the work done by pulling force not only provides an increase in kinetic energy but also an increase in potential energy!

So it is: w=mgh+1 2mv2

In fact, w = 1 2mv2

That is, WF + WG = 1 2MV2

wf-mgh=1/2mv2

Gravity has active work in it and is negative work.

It would be easier to analyze the man and the ship as a whole, because the displacement of the ship is the displacement of the position, and do not be confused by what the ship is connected to. First of all, this situation ignores the friction between the boat and the water.

Let the total mass of man and ship be m, a, the system composed of man and ship is f, the displacement is s, the work is f*s, and the man and ship obtain velocity v0, f*s=(m*v0 2) 2-0

b. The combined external force of the system composed of man and ship is 2F (the person is pulled to the right f, and the ship is also subjected to the right f), and the displacement is also s, 2f*s=(m*v 2) 2-0, and the solution is v = root number 2*v0

c. Same as a, the system is subjected to the resultant external force f, and the displacement is s, so the velocity is v0

Ignore all resistance:

The first case: fs=1 2*mv0 2

In the second case, the boat has gone away, and the rope in the person's hand has changed by 2s, so the work done is 2fs, v = v0 twice the root number

In the third case, according to the conservation of momentum, the C boat moves in opposite directions, so the final velocity of the two ships is opposite in the direction and equal in magnitude.

At this time, the work done by man should be 4fs, but at this time there are two ships moving, so 4fs = 1 2*mv 2+1 2*mv 2, and the solution v is also equal to v0 which is twice the root number

The first case: fs=1 2*mv0 2

In the second case, the boat has gone away, and the rope in the person's hand has changed by 2s, so the work done is 2fs, v = v0 twice the root number

In the third case, according to the conservation of momentum, since the masses of the 2 ships are equal, the final velocity of the two ships is opposite in the direction and equal in magnitude.

The rope moves for 2s relative to the person, and the two boats are moving, so 2fs=1 2*mv 2+1 2*mv 2, and the solution v is equal to v0

It can also be solved according to the acceleration s=1 2at 2

In 2, the acceleration is equal to 2a, so t=t root number2 v= at = 2at root number 2 = v0 of 2 times the root number

In the 3rd the acceleration is equal to a, so v=at=at=v0

24.As shown in the figure, AOB is a slide model in the playground, which is located in the vertical plane, connected by two circles with a radius of R1 4, and their centers O1 and O2 are connected in the same vertical line as the connection point O of the two arcs. O2B is along the surface of the pool, and O2 and B are on the same level.

A small slide with a mass of m can be slid down from a standstill at any position in the arc AO, regardless of friction.

1) Assuming that the small slider slides down from point A, find the pressure on point O when the small slider slides to point O;

2) What is the distance from the water point to O2 for the small slider that can get out of the slideway at point O;

3) If the arc length of the small slider is equal in the process of sliding from the beginning to the time it leaves the slideway, then where should the small slider be on the arc AO when it starts to slide down (expressed by the trigonometric value of the angle between the line to the O1 point and the vertical line).

Solution: (1) mgr = mv2 1 point.

FN—MG=MV2 R 1 point.

Synopid: fn=3mg 1 point.

From Newton's third law: the pressure is 3mg, and the direction is straight down. 1 point.

2) The speed of the slider sliding from point A to point O is , and the minimum speed of the deorbit is V1, then there is:

mg = MV12 R, 1 point.

Score: v1 = 1 point.

r= gt2 1 point.

x=vot 1 point.

Synopid: R x 2r 1 point.

The ball is thrown vertically, the maximum height of the rise is h, the resistance is constant, the ground is zero potential energy surface, when it rises to the height h above the ground, the kinetic energy of the ball is 2 times the potential energy, and when it falls to the ground height h, the potential energy of the ball is 2 times the kinetic energy, then h is equal to.

Pure hand-hitting.

Set the coefficient of friction, the mass is 2m, m

Start status. f=μ（2m+m)g=3μmg...1) Disconnected, B glided S1

mgs1=(1/2)mv² .2) F removed A and glided S1

2mg)s2 = (1 2)(2m)v + f + mg)s substitution (1), (2).

2mg)s2=2μmgs1 +4μmgss2=s1 +2s ..3) ab is far apart.

s2-s1+s+l

3s +l

As for the functional transformation in the process of shot put, it is the conversion of gravitational potential energy into kinetic energy, but it has nothing to do with athletes. This 8 8mv 8 in ,..Instead of gravity doing it, you can understand it this way, the kinetic energy theorem says that the combined external force, then part of the force exerted by the athlete during the throw will be gravity.

Let the friction factor U, the mass be 2m, m

Start f=u(2m+m)=3umg, after the two are disconnected from l, to b,1 2mv2=umgs

For a,1 2MV2=fs-2UMGS

The ball leaves the hand with kinetic energy = 1 2mv*v, which is the work done by the human hand on the ball.

Another 1 2mV*V=mGH+FH

So choose AD

Assuming that the work done by the hand is w, then w = mv2 = fh + mgh

A and D, the hand is off the ball and cannot do work.

The kinetic energy theorem is not columnar like this, the left side of the equal sign is the kinetic energy variable, and the right side is the work w on it. Thereinto.

h=1m*sin30=；

So, the ascending phase:;

Descending Phase:;

In this way, the solution of the cube is vt=2m s

Equation 2 is incorrect, in this case mgh-fs= or solution:

m g h - 1 2 m v0 2 = -u The work done by friction during the ascent process is u

1 2 m v0 2 - 2 u = 1 2 m v 2 The work done by friction in the whole process is 2u

Substituting yields v=2

Up: -mgh-fs=-mv0 2

Down: mgh-fs=mv 2, I see that the relationship between your columns is very awkward, I don't want to make any sense, you don't understand the positive and negative of the work.

To give you an idea, let the initial state potential energy be zero, and there is only friction in the whole process, and the ascent process, the kinetic energy ek0 is converted into potential energy and overcomes friction wf and the final kinetic energy is ek1=ek1-2wf