It s a bit of a difficult question, it s a high school physics question, and it requires a process a

The mass of the object at rest on the horizontal ground is 2kg, and it starts to move under the impetus of the horizontal constant force f, and its velocity reaches 4m s at the end of 4s, f-f=ma1, v=a1t1

Remove f, and stop after 6s object, f=ma2, 0=v-a2t2

Solve the system of equations to get f=

Let the dynamic friction factor be u

From "its velocity reaches 4m s, f is removed at this time, and the object stops after 6s", ugt2=mv can be obtained, and u=m*v g t2=2 15 can be obtained

(F-UMG)*t1 m=mv can be obtained by "starting the motion under the impetus of a horizontal constant force f, and its velocity reaches 4m s at the end of 4s, at which point f is removed".

So f=20 3(n).

v=v0+at so 4=0+a*4 so a=1

So f-f(friction)=ma and because f=u(coefficient of dynamic friction)mg, f-u*20=2

After removing f, a =ug a t =4 so a =2 3 so u = 2 30 so f = 10 3

I'm also so difficult, I want to ask a lot of questions in junior high school, and a big question in high school.

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Select d from the diagram to see that the force analysis of the three balls is similar, so only one can be analyzed, refer to a small ball, f bounce = mg,

At the moment when the hanging basket rope is cut, the position relationship of ABC does not change, B and C are still supported on A, A and C are not separated, the distance between BC remains unchanged, and the spring elasticity does not change. b by gravity and spring support, balanced, acceleration is 0; c is subject to gravity, the elastic force of the spring (equal to the gravitational force of b), and the supporting force of a. For a certain period of time, C is pressed on A by the spring elastic force, and moves with A, and the acceleration is the same, set to A, and the AC force is N.

A is subject to gravity and n, mg ten n = ma, 6 mg ten 2 mg-n = 2 mA2 mg ten 2n

3n=6mg，n=2mg

a=3g。

Buses depart from bus stations every T0, indicating that the distance between each two buses is the same. So the problem translates into people and cars in the same direction for 3 minutes, and the reverse is 5 minutes. Let the speed of the vehicle v1 and the speed of the person v2, the distance between the two cars is s, and the distance between the two cars is s, and the distance between the two cars is s

v1+v2）*3=s

v1 -v2）*5=s

Here are the graphs and formulas, and the answer is up to you.

(1) Let the velocity of the small slider be v when it can reach the q point, then qe+mg=mv r (the circular motion can reach the highest point condition).

In the process of the slider moving from the beginning to the q point, the kinetic energy theorem has:

(mg+qe)x-(mg+qe)·2r=½mv²-½mvo² (x=

Simultaneous solution: vo = 7m s

2) Let the velocity of the slider when it reaches point P be v1, then the process from the beginning of motion to point P, the kinetic energy theorem:

(mg+qe)x-(mg+qe)r= mv1 - mvo at point p, the circular motion obtains: the support force of the orbit on the slider f=mv1 r simultaneous solution: f=

By Newton's third law: pressure is.

1.With the law of conservation of energy, mgh + 1 2mv 2 = 1 2mv'2, v=5m s.

2.I forgot about the formula, sorry, hehe.

square + mgh = squared.

The square of MV r+mg=f

square + mgh = squared.

Haha Help you learn force analysis (1) Determine the research object : cylinder (2 Analyze gravity ) Any object is subject to gravity Size: mg Direction:

Vertical downward point of action: center of gravity (three to analyze the elastic force by looking at who the object to be analyzed is in contact with) The cylinder is in contact with the air and the gas in the cylinder is in contact with the piston * The air has an upward pressure on it f1 * The gas has a downward pressure on it f2 * The net force of the piston on it is zero (4. Analyze the friction between the object squeezed by the object under study ) Regardless of friction.

Cylinder force balance f1=f2+mg

You ignore the atmospheric pressure, according to the method I said, the force analysis will not be more force, and the force will not be less, take the method I gave you seriously, I hope it will help you.

According to the title, the meaning, the pressure of the gas in the cylinder is less than the atmospheric pressure, and when you lift it, the pressure of the cylinder by the outside atmospheric pressure is upward. Self gravity downward. All are MG, and the force is balanced.

A perfect score of 120 points on this question should not be able to score more than 90 points, after grabbing the piston and lifting the cylinder, the gas pressure inside the cylinder is less than the atmospheric pressure outside. According to the force analysis of the cylinder, the sum of the gravitational force it is subjected to and the pressure of the internal gas on it is equal to the upward pressure of the outside atmosphere. In the vertical direction, the outside atmospheric pressure has an upward force on the cylinder instead of a downward force.

You should be the second question, no.

This one mainly analyzes the direction of friction.

When the conveyor belt is not moving, the friction is upward.

When teleporting counterclockwise.

If the velocity of the conveyor belt is greater than the velocity of the object, then the relative direction of motion of the object is upward, and the friction force is downward, and the natural acceleration is greater than the original, and the time is reduced.

If the speed of the conveyor belt is less than the speed of the object, the friction is upward (the same magnitude as when the conveyor belt is stationary and clockwise), and the time does not change.

Equal velocity is only possible for a split second, and the object is always accelerating.