About the actual operation of Newton s second law! Urgently need thanks. 5

There are three forces in the direction of motion along the inclined plane:

1. The component of the f force along the inclined plane = f * cos (s)2, the component of the object's gravity mg along the inclined plane down = mg*sin (s)3, the dynamic frictional resistance (downward) formed by the partial force of the f force and the vertical inclined plane of the object's gravity mg

mg *cos (s) + f * sin (s)]*df net force = f * cos( s) -mg*sin( s) -mg *cos( s) +f* sin( s)]*d = ma

f = m * a + g * sin( s) +cos(s) *d))/(cos(s) -sin( s) *d)

Splits along the inclined plane into perpendicular to the inclined plane and parallel to the inclined plane direction.

Perpendicular to the inclined plane fsin +mgcos =n, along the parallel to the inclined plane fcos -(mgsin +f)=fcos -(mgsin +d*n).

fcos -(mgsin +d*(fsin +mgcos) =ma, I don't need to say it later.

Trigonometric Identity Deformation Formula: Trigonometric function of the sum and difference of two angles: cos( =cos 2.Newton's second law of motion: f plus ma or a f v ma { is determined by the resultant external force, and the direction of the resultant external force.

1 Q: s=a*t 2 2 get the acceleration a m s 2

2 Q: f f ma 2 f 8* to get friction f calv.

3 Q: At the end of 10 seconds when the wooden block has a pulling force, the speed is v a*t1

In the absence of tensile force, f ma1 gets a1 m s 2

If the gliding distance is set to s1, then 0 v 2 2*a1*s1 will get s1 40 meters.

Answers to questions supplemented:

In the analysis of the sliding process, the force is known as mgsin37 degrees, mgcos37 degrees, ma1

Get a1 8m s 2

The total time of its upward sliding is set to t1, then t1 1 second is obtained from 0 v0 a1*t1, and the total distance of the upward glide is s1, then there is s1 a1*t1 2 2 4 meters (reverse thinking).

In addition, since mgsin37 degrees and mgcos37 degrees (this relationship can be calculated by substituting the value), the block will continue to slide down after reaching the highest point.

When moving downward: mgsin37 degrees, mgcos37 degrees, ma2, a2 4m s2

In t2=t total t1 2 1 1 seconds, the glide distance is s2 a2*t2 2 2 2 2 meters, and the displacement s1 s2 4 2 2 meters in the 2 seconds from the beginning of the upward movement to the end of 2 seconds, and the displacement direction is along the inclined plane.

The velocity v a2*t2 4m s, the direction is downward along the inclined plane.

Choose. CA mass is a property of matter.

It does not change due to force.

b should be said when the quality is certain.

Acceleration is directly proportional to the force.

Because of force, there is acceleration.

The acceleration changes because of the change in force.

Instead of the force changing because of the change in acceleration.

Besides, what he said was not right either.

It needs to be quantitative.

You can only judge if you know that it does not change.

dWhen the mass increases.

If the force also increases.

Acceleration may not change.

It may also be enlarged.

The initial velocity can be calculated:

h=70cm=

a=g=10m/s^2.In the air, the acceleration is -g, and the question gives the acceleration before the take-off off leaves the ground, which accelerates it from 0 to the initial velocity.

The ascent process can be calculated with the reverse equivalent:

v^2=2gh

v=√2×10×

If you want to calculate the support force, it's simpler: acceleration a=(fn-g) m => fn=mA+g=ma+mg=4010m=401g

Therefore, fn = 401g, that is, the ground it bears when it takes off is 401 times more supportive than its body weight.