High School Physics An object with a mass of 10kg moves at a constant speed along a horizontal plane

1) When moving at a constant speed, there are two forces equilibrium f friction = f tensile force = 20n, and f friction = fn is obtained by the formula of sliding friction

So =20 100=

2) From Newton's second law.

f Pull - f Friction = mA

30-20=10a

a=1m/s^2

3) After removing the tensile force, there is only sliding friction, and the original direction of motion is positive, which is obtained by Newton's second law.

fFriction = ma'

a'=-20/10=-2m/s^2

A negative sign indicates that the direction of acceleration and the direction of motion are opposite.

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1) g = 10kg 10n kg = 100n (see if there is g on the test paper to take 10n kg, if not, use it.)

f g = A: The dynamic friction factor is.

2）f=30n-20n=10n

a=f/m=1m/s²

Answer: The magnitude of acceleration is 1m s

3）a=f/m=2m/s²

Answer: The acceleration magnitude is 2m s

1) Moving at a constant speed at 20N, the frictional force is equal to the tensile force. f pull = 20n = f g = 20 10 * 10 =

2) a=f m = (30-20) 10 = 1m s 2

3) a=f m=20 10=2m s 2

20n horizontal tension constant velocity, so mg=20n, so =20 g (unitless).

The dynamic friction force of the object on the horizontal plane is 20N, so with a force of 30N, the acceleration a=20Nm=1m s2

After the pulling force is removed, the object only acts on the kinetic frictional force, so the acceleration a2=20n m=2m s2

Find the acceleration v=a*t, a=v t=6 2=2m s 2f combined=m*a=2*2=4n

f = f pull - f friction = 4n

Then f friction = f pull - f fit = 6n (opposite to the direction of motion) after the external force is removed.

Acceleration a=f friction m=-6 2=-3m s 2 (do uniform deceleration motion) v=v0-at=0 calculate the motion time, v0=6m s t=2s then s=v0t-at 2 2= 6*2-3*4 2=6m

f=10n because the motion starts from rest v3s=at=6m s a1=2m s 2=f f f 1=4n f 1=f-f friction f = 6n so the ground friction during the motion of the object is 6n f 2=f friction = 6n after the external force is removed, f 2=f friction = 6n a2=f 2 m = 3m s 2 at the end of the 3s The velocity of the object reaches 6m s v = 0 So t deceleration = v3s a2=2s When decelerating, do uniform deceleration linear motion v average = 1 2v3s = 3m s Therefore, the distance that the object can continue to slide after the external force is removed is x=v, the average *t, and the deceleration=6m

Xuanshangf bounty points are too much, the object does a uniform acceleration motion in the horizontal direction, and the force analysis of the object:

Gravity, support, tension and friction, the resultant force is the pulling force minus the frictional force is 6n According to Newton's second law, the acceleration of an object can be obtained.

a=f m = 2m s 2

by v=v0+at(v0=0).

The velocity of the object at the end of 3s is 6m s

First, find the acceleration method: f=ma (f=10-4=6n, m=3kg) The object moves in a uniform accelerated linear motion with zero initial velocity.

Therefore, v=at (v is the instantaneous velocity at the end of 3 seconds, t is 3 seconds); Because the question is about the amount of velocity, you don't need to consider the direction of velocity. Thank you.

The resultant force is 10-4=6nThen the acceleration is a=6 3=2m s squared, v=at=2*3=6m s

First, find the external force f = 10-4 = 6n f = ma a = f m =6 3=2m s v3=at=2 3=6m s

Moving at a constant speed, it is subject to a balanced force, so the frictional force f=5n

After removing the push-source modulus force, f=ma

a=f/m=5/2=

The object decelerates its motion, v0 = 2as

s=v0 2a=5 hail paratar 2*

It is done along the horizontal plane under the action of horizontal tension of the size of 5n.

Uniform linear motion.

It shows that the friction of the silver group on the ground is 5N

1. Leakage f-f=ma

9-5 = 2a Pulse Pede a = 2 m s 2

2、f'-f=m*3a

f'-5=12 gives f'=17n

1. The work done by the pulling force is the same, because W=Fs, F is the same, S is the same 2, the work done by each force is different.

w total = wf + wf = the work done by friction plus the work done by the pulling force has no friction on a smooth horizontal plane, so the work done by friction is less, so the total work of the two is not the same.

Solution: w fs, both the pull force and the distance are the same.

w＝15n×

The object moves on the horizontal plane, the pulling force does positive work on the object, the frictional force does negative work on the object (or the object overcomes the frictional force to do the work), w total fs fs, the object moves on a smooth plane, the frictional force does the work is zero, then, the total work: w total fs

The object moves on a rough surface, frictional force: f mg

Total work: w total fs fs 15n

The work done by the pulling force is the same, the first one moves on the horizontal plane, and the work done by the pulling force is all converted into its kinetic energy.

The second type of motion on a rough surface, the work done by the pulling force is equal to the kinetic energy of the object plus the work done by the frictional force.

It is the same in (1) The work done by the tensile force is fx, f: the magnitude of the force, x: the distance the force moves (2) The position of the object changes is the same, so the total work of the object is also the same.

1 All f=

acceleration section acceleration a=(f-f) m=(10-2) 2=4m s 2 acceleration section displacement s1=(1 2)at 2=

Acceleration of the deceleration section a'=-f m=-2 2=-1m s 2 The initial velocity of the deceleration section v = the final velocity of the acceleration section = at=4*1=4m s The displacement of the deceleration section is set to s2

2a'*s2=0-4^2=-16

s2=-16 (2a)=-16 [2*(-1)]=8m, total displacement, s=s1+s2=2+8=10m

In 0-2s, the acceleration of the object is (10-5) 2=, then the net force of the object is 5n, and the acceleration of the object in 2-6s is 10 4=, at this time there is no tension, there is only friction in the horizontal direction, so the friction force is 5n

So within 0-2s, call Hu f-f=ma, then f=10n handsome wolf hunting and troubleshooting Ying] team will answer for you in the bend.

The second half gives you the calculation of friction, and the reverse acceleration is, and the friction force is 5n.

In the first half of the blockage, the positive acceleration is the size of the object, and the resultant force is 5N, and the tensile force-friction force = the resultant force is 10

By digging out the balance of the flat air dispersion at a constant velocity, the friction force ff-f1=0 of Douchun can be obtained

f=f1=20n

f-f=ma

f=f+ma=20+10*1=30n

a=1.Therefore, the combined external force is 10n. So the pulling force is 30N

a*t=16

a*t*t/2=16

The simultaneous equation is solved to obtain the acceleration time t=2 acceleration a=8 combined external force = 100 - friction force = a*m = 8*10 The solution is as follows: friction force = 100-80 = 20

After the withdrawal, the acceleration is -20 10=-2, and the speed is reduced to 0 after 16 2=8 seconds, and then it is no longer moving.

Displacement = 16 * 8 - 2 * 8 * 8 2 = 64

Solution: (1) The motion time of the Dan cluster is t, and the acceleration is a

Because at=16,1 2 at 2=16, so t=2(s) so a=8m to tell the late letter s 2

So if there is f-f=am, then f=20n

2) Because f=20n, so a1=f m=2m s 2 because sock wheel v1=16m s, a1=2m s 2, t1=10s, so s=v1 t1-1 2 a1 t1 2=60m