A high school object problem, a high school physics problem

a . c .The object slides down an inclined plane. Then only g does work on the object. The oblique does not do work on the object.

Therefore a is correct. According to the kinetic energy theorem, the algebraic sum of the work done is the amount of change in kinetic energy. So c is correct.

ad correct. B, the inclined plane and the ball are always vertical, do not attack, because the inclined plane is protected on the smooth horizontal plane, there is no friction, so the following will also slide, it is the work done by the ball on the inclined plane, part of which is converted into the kinetic energy of the inclined plane, so the kinetic energy is less than mgh when it reaches the ground

aBecause there is no friction, the object does not do work on the inclined plane.

c Because only gravity does the work, the change in kinetic energy caused is the same as the gravitational potential energy.

According to simultaneity, the correspondent used 40s, and the team also used 40s. Then V team = 200 40 = 5m sThe communicator walked to the point where the team advanced 200m, and also returned 40m, and the communicator walked 200+40*2=

This kind of question is considered in isolation, starting with your question:

Isolate the female athlete, the force is tensile force t and gravity mg, the resultant force f of these two forces should point to the center of the circle that rotates horizontally, that is, perpendicular to the axis of rotation, and the trigonometric relationship mg=tsin, t=mg sin

It is more convenient to find the angular quantity of rotation naturally, and the resultant force obtained by f=mg tan is obtained by the centripetal force formula f=mrw 2, and w= [g (rtan)] period t=2 w is obtained.

1) For columns A and B, for columns A there are:

t – magsinθ-f=maa1 ; fn=magcosθ ;f=μfn

For b there is: mbg-t=mbA1

Combine the above to get: mbg-mba - magsin -magcos =ma1 (1).

For the movement of B, there are: V2=2A1H (2) A1=4M S2 is obtained by substituting data from (1) (2), =(2)After B lands, the rope relaxes and no longer has the effect of pulling force T on A, and A has Magsin +F=MAA2; fn=magcosθ ;f= fn is arranged: a2=, the direction is downward along the inclined plane, so a continues to do uniform deceleration motion along the inclined upward direction, and the displacement is x=v2 2a2=4 13m.

The maximum distance for an object to slide along an inclined plane is s=x+h=21 26m.