Newton s second law in physics, Newton s second law in physics

1.Solution: v=8m s, t=10s

a=v/t=

2.Solution: f=ma=2kg*

f=6n-f=

f=At this time, a=f m=

v0=,a=

The magnitude of the velocity of the object at the end of v1=5s is 1m s

This is the specific process, what will not be asked!!

After analysis, this object is subjected to tension and friction, so the resultant force is the pulling force minus the frictional force. According to the velocity at the end of 10s, find the acceleration = meter per square second, the resultant force = ma = , so the friction force is equal to f = f - ma = , remove the tensile force only by the friction force, the direction is backward, so do a uniform acceleration linear motion, acceleration = friction divided by the mass equal to the meter per square second, t = v a = 8

v=a1ta1=v/t=8/10=

f-f=ma1

f=f-ma1=6-2*

If the external force is withdrawn at the end of 4s:

v=a1t=

Deceleration after removal of external force: a2=f m=

t2=t-t1=5-4=1s, that is, the end of 5 seconds is equivalent to the end of 1 second after deceleration.

v=v1-a2t2=

You ask the head of the regiment, I will only be in the second year of junior high school.

1) Move for uniform acceleration.

a=v/t=8/10= m/s2

2) Second question? What do you mean?

1。Wrong, the moment the car starts, the external force is not zero, but there is no speed.

2。True, acceleration is only related to the combined external force and mass, and the greater the combined external force of the same object, the greater the acceleration.

3。Wrong, the car brakes, the external force is backward, but the car is forward.

Example of an accelerated motion of an object with an external force becoming smaller:

I'll make one, a spaceship, not subject to any gravitational pull, now going forward with a rocket booster, the thrust is smaller, but the ship is accelerating.

1. Incorrect, when the object has a certain initial velocity, and now the external force is not zero, but the direction is opposite to the direction of motion, and the object does deceleration, then there must be a moment when the velocity of the motion is 0, and then it begins to accelerate in the opposite direction.

2. Correct. 3.Incorrect, when the object is decelerating, the direction of the external force is opposite to the direction of the velocity. Also, are there any examples of an object accelerating when the combined external force becomes smaller?

Answer: As long as there is an external force, it must be an accelerated movement, hehe!

1.False, the velocity may be zero, for example: the starting point of a free fall motion (vo = 0, but the resultant external force is equal to the gravitational force).

2.Correct (according to Niu Ge's law).

3.Error, example: a box sliding on the desktop. The resultant external force experienced by the box is equal to the frictional force, which is opposite to the direction of his motion.

The resultant external force becomes smaller, but the acceleration becomes smaller, but still doing the acceleration motion is an example: acceleration is because the direction of velocity is the same as the direction of acceleration. When an object falls, its resultant force is gravity downward, and when we apply a slight upward force (this force is small, less than gravity), its resultant force is still downward, but decreases.

The object still has to accelerate ......

1.Error, the velocity can also be zero, such as a stationary object 2That's right.

3.False, the direction of the resultant external force should be the direction of acceleration, not the direction of the object's motion The resultant external force becomes smaller, but the acceleration becomes smaller, but it is entirely possible to do the accelerated motion.

1. Incorrect, when an object has just received an external force, we can understand that it has a tendency to move and is at rest.

2. Correct. 3. Incorrect, for example, if you throw a ping pong ball vertically upward, the table tennis ball is affected by gravity and the external force is downward, but when the ball reaches the highest point, the direction of movement is upward.

4. If the resultant external force you mentioned becomes smaller and disappears, the acceleration of the object will gradually decrease until it moves at a uniform speed.

1.Incorrect! The velocity is 0 and the acceleration is g at the highest point of the object thrown vertically.

2.That's right. f=ma, when the resultant external force increases, the acceleration also increases.

3.Incorrect! When the direction of the combined external force and the direction of velocity are the same, the object accelerates;

When the direction of the combined external force is opposite to the direction of velocity, the object is in deceleration motion.

Are there any examples of an object accelerating when the combined external force becomes smaller? The variable acceleration linear motion with reduced acceleration and the external force becomes smaller, the object can accelerate the motion, and it can also do deceleration motion!

1.errors, the speed may also be zero, such as free-fall motion. 2.

Correct 3False, the direction of the resultant external force should be the direction of acceleration, not the direction of the object's motion, such as flat throwing motion. The combined external force is gravity, and the direction is vertical downward, but the direction of motion is the tangent direction of each point on the curve, which is changing all the time.

It is entirely possible that the combined external force becomes smaller, but the acceleration becomes smaller, but it is still accelerated. For example, if the initial velocity of the object is to the right, the resultant external force is to the right, but the magnitude becomes smaller.

There is no direct correspondence between the external force and the motion of the object. With the acceleration definition a=(v2-v1) t, it can be seen that the direction of the resultant external force and the direction of velocity change are the same. There is no direct correspondence with the direction and magnitude of velocity.

At the moment when the rope is cut, the spring has no time to deform, so the elastic force remains unchanged, so the acceleration of B is zero, so the elastic force of B must be balanced with gravity, so it is MBG. For A, it is pulled by gravity and spring, so it is (ma+mb)g.

The holistic approach and the segregation approach are discussed separately.

b is only affected by gravity and elastic force. and in static equilibrium.

For b: Balanced by force mbg=t

Pair a: Balanced by force mag+t=f rope.

Due to the abrupt nature of the rope and the slow-changing nature of the spring.

After cutting the rope. For a is only affected by the downward mag+t=mag+mbg, so the acceleration is (ma+mb)g ma

For b, the force is still balanced. a=o

ascending process a=(mg+f) m=12m s 2, downward;

Descending process a=(mg-f) m=8ms2, downward.

The velocity is 0 to v on an inclined plane, and then to 0

v=a1*t1=a2*t2;(Regardless of the positive or negative acceleration) drawing shows that the pressure of the object on the inclined plane is mgsin37, and the friction force = mgsin37 , along the inclined plane downward.

t=4s，v=a2t=4μg，s2=1/2（a2t²）=8μg；

s=s1+s2=16gμ²/cos37º-μsin37º)+8μg

The upper and lower subscripts on the inclined plane are 1, the upper and lower superscripts on the plane are 2, and the velocity at the moment when the inclined plane slides out is v1, and * denotes multiplication.

v1 = a2 * 4s = * g *4s = g * 1sa1 = g * sin37° *g * cos37°t = v1 / a1

s = * v1 * t

Using the law of momentum to analyze, it is very simple, I can't calculate the distance, because the conditions are not enough, but time can be calculated: and you told me that the kinetic friction factor is a superfluous condition, as long as you find that the kinetic friction factor of the inclined plane and the plane are the same. The timing is:

4/(sin37-cos37).

The kinetic energy theorem can be used.

Is there a quality of wood???

d, because the object should have the same acceleration as ***, the gravity of the object minus the support force should produce a downward acceleration, and the support force also reflects the indication of the pallet, so the support force should be light rain gravity, so it should be smaller.

aExcluded, the mass will not change due to the change of the state of motion.

b excludes, the object is always equal to mg by gravity

Row C, which is the pressure of the object on the scale.

d pair, as can be seen from Newton's third law, the action force is equal to the reaction force, where the object accelerates and falls, the resultant force decreases, and the supporting force decreases.

Therefore, the pressure is reduced.

The object has advanced 25m from rest after 5s, and according to the formula s=1 2*at 2, the acceleration is a=2m s 2

According to Newton's second law f-f=ma, f=f-ma=25-5*2=15nu=, so the frictional force f=umg=, it can be seen that the tensile force is f=12n withdrawing f, the frictional force produces acceleration a=f m=12 3=4m s 2, the initial velocity v0=8, the final velocity vt=0, according to the formula vt 2-v0 2=2as, s=8*8 8=8m

1.The object moves in a straight line with uniform acceleration.

Niu 2 f-f = ma

and x=(1 2)at 2

The acceleration of the motion of the object is solved a=

The frictional force of the bottom on the object f=

2.Wooden blocks are transported in a straight line at a uniform speed.

f=umg remove f, and the wooden block is transported in a straight line with uniform deceleration.

Niu Er, -umg=ma

0-v0 2=2ax

The block can also glide on the horizontal plane x=

1 x=v0t+1 2at 2 1 2a 5 2=25 a=2

f-f=ma f=15n

2 f=f= v0 2-v end 2=2ax Substituting data yields 8 2-0=2(12 3)x

x = 64 8 = 8 meters.

First of all, f (combined external force) = ma, and then f (combined external force) is the resultant force of tensile force and elastic force and gravity, the tensile force is f, the elastic force is kx, and the gravity force is g, where the - sign is related to the selected direction, only a does uniform acceleration motion in the process, ma is a fixed value, so f = fixed value + kx-g, and x is equal to at, so f is a linear function, the tensile force gradually increases and the initial value of the tensile force is not 0 [otherwise it will not move], over, no thanks!

The upper block is subjected to the upward pull and the downward pull of the spring, and the block accelerates upward, so the upper pull f minus the lower spring pull kx is equal to the upward net force ma.

Because it is originally in equilibrium, the force f starts from nothing and starts from the origin, and the displacement also starts from 0.

Isn't the answer A? Since he does a uniform acceleration motion, his combined external force is 0, how can f be 0 at the beginning.