18 An object with a mass of 10 kg under the horizontal thrust of F 200N

Let's talk about the knowledge of impulse! ~！It's that simple! 1. According to the title, the friction force acts for 2 seconds, and the pulling force acts for 2 seconds.

Let's study the forces in the direction parallel to the inclined plane.

There's f*'*2 f'=f*cos37 Isn't that the friction out! 1. Isn't it easy to find the coefficient of friction? ~！

The total displacement is calculated in two parts: acceleration up and acceleration downward! Because I don't have any paper in my hand, it's inconvenient to calculate for you! 1. The room in terms of calculation is troublesome to the landlord.

Before withdrawal, n1 = g*cos37 + f * sin37 = 200 f1 = n1* = 200

m*a1=f*cos37-f1-g*sin37=100-f1=100-200μ

v=a1*t=2a1=

After withdrawal, n2 = g * cos37 = 80

f2=n2*μ=80μ

m*a2=f2+g*sin37=f2+60=80μ+60100-200μ)/(80μ+60)= μ= a1=5 a2=8s=

The force analysis of the object is carried out, and after the force is removed, f=umgcos37=maa=ugcos37=8u

When F acts on an object, there is FCOS37-U(mgcos37+FSN37)=MA1

a1=16-20u

Let the velocity of the object be v when the force is just removed

then v=a1*2=a*

That is, 8u* solution u=

a1= a= v=

s=v^2/(2*a1)+v^2/(2*a)=

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

There is a kinematic formula vt-v0=at to get 6-0=3a a=2m s2

The positive pressure n is equal to the weight of the object n=g=mg=2*

According to Newton's second law, f-un=ma, then the dynamic friction factor u=(f-ma) n=(10-4).

Let the dynamic friction factor between the inclined planes be U, the acceleration in 2s is A1, and the acceleration after removing F is A2,1) In the first 2S, the object is acted on by four forces, horizontal F, vertical downward gravity g=mg, support force N, and the oblique downward friction force of Mola oblique downward f=un, decompose F,G into two directions along the inclined plane and the vertical inclined plane, the direction of the vertical and straight inclined plane is: N=GCO37°+FSN37°, and there are FCOS37°-GSN37°- along the direction of the inclined plane- un=ma1, A1=10-20U is obtained by the two formulas, and the velocity at the end of 2S should be V=A1*2S20-40U, and after removing F, the object is still sliding upward, and the direction of friction remains unchanged, at this time A2=GSN37°+UGCOs37°=6+8U, and from this process, V=A2*The above two Vs are at the same time, so the magnitude is equal, and U=1 4= is obtained from the equation

2) The front and rear ends are moving at a uniform speed, and the displacement formula can be calculated by adding the displacement of the two stages. The displacement in the first 2s is s1=a1*(2s) 2 2=10m, and the displacement of the rear section is s2=a2*(the total displacement of the bridge body is s=s1+s2=

First draw a drawing, find a rental search for various forces, gravity, friction, tension, establish your own coordinate system, flat and miss the slope of the manuscript line, and then apply the formula. Physical mechanics is very important, and you should practice more.

Hope that helps.

The kinetic friction factor between the object and the ground is known, and m=1kg, f=10n, so the frictional force f = = 2 n

The resultant force of the tensile force and the frictional force f = f-f = 8 nThe acceleration of the motion of the object a = f = m = 8 m s2, so the velocity of the object at the end of 2s v = at = 16 m s, so the velocity at the end of 2s is 16 m s

f=ufn=umg=

f-pull = f-pull-f=10n-2n=8n

a=f m = 8 1m s=8m s

v=at=8*2m/s=16m/s

To sum up, the speed is 16m s

Friction force f = = 2 n

The resultant force of the tensile force and the frictional force f = f-f = 8 nThe acceleration of the motion of the object a = f m = 8 m s2 The velocity of the object at the end of 2s v = at = 16 m s

The work of the pulling force wf=fs=while moving on a smooth horizontal plane

Other forces (gravity, support) do not do work.

While moving the same distance on a rough horizontal plane, the work of the pulling force wf=fs does not change.

Gravity and support still do not work.

Friction does negative work wf=fscos180°=- mgs

w=fs

Because f is equal to 15n s, it is equal.

Hello! Because it is moving at a uniform velocity, the frictional force f=f=20n

When the thrust is withdrawn, the object moves with the speed of the orange branch a=f m=20 4=5m s 2, and the Ghenbu Wooki can be obtained according to vt 2-v0 2=2as: s=8 2 2*5=

It is easy to 1) by 2as=v2 2a* a=2m s 2(2)f- mg=ma=

Just choose me.