A physical problem about force, solved

First of all, I will give you a diagram of the force analysis of an object

The object is subjected to friction, support, vertical gravity, and tensile force f along the inclined plane, then in such a force, the object starts to move from rest and moves 4m in 2s, then the object is moving with uniform acceleration.

After analyzing the motion state, we can calculate the acceleration along the inclined plane.

1) It can be calculated from the force analysis that the supporting force fn=mgcos37°=80n on the object, and the sliding friction on the motion of the object, f=fn* =mgcos37° =80, then the resultant external force f' of the object along the inclined plane direction is equal to the tensile force minus the component force and frictional force of gravity in the inclined plane direction, i.e.

f'=f-mgsin 37°-f

100-10*10*, from the question "the displacement of the object in 2 s is 4 m" and then according to the kinematic formula, there is 1 2at*t=s, substituting the number to calculate the acceleration a=2m s2, there is Newton's second law to know, f'=ma, substituting the acceleration value, and f can be calculated'=20n, which is calculated by substitution, = .

2) After removing the force, the resultant force of the object is down along the inclined plane, the acceleration is a=g*(sin 37°+cos 37° )=8 m s2, at the end of 2s, the velocity of the object is v=4m s, then after the force is removed, the object does a uniform deceleration motion and knows that the velocity is zero, and the time taken is t=v a=, so when the velocity of the object is reduced to 0, but because mgsin 37°-mgcos 37° is greater than 0, so in the absence of tensile force f, the object will slide along the inclined plane, At this point, the direction of the sliding friction experienced by the object is upward along the inclined face, then there is ma'=mgsin 37°-mgcos 37° to get the acceleration a when the object slides down'=4m s2 for 1 second, at which point the velocity of the object v'=4*1=4m s

3) The displacement is a vector quantity, and the distance is a scalar quantity, so here the motion of the object is divided into three stages, the first stage, 2s from the stationary starting motion, the displacement s1=4m, the second stage, after removing the force, the object does a uniform deceleration motion along the inclined surface, the displacement s2=1m, the third stage, the object does a uniform acceleration motion along the inclined surface, the displacement s3=-2m, so the displacement of the object in the S from rest starts s=s1+s2+s3=4+1-2m=3m.

The distance is l=4+1+2m=7m.

Got it? <>

1. The object is accelerated by stationary motion, and S1=4M, A1*T1 2=S, A1=2M S, V1=4M S, the object is subjected to an oblique upward pull force f along the inclined plane direction during the motion, the oblique downward gravity component mgsin = 60N, and the downward friction force of the inclined object mgcos. m*a1=f-mgsinθ-μmgcosθ

Obtain =2, remove the force f, and the force f, the force on the object along the inclined plane is f'=mgsin = 60n, friction f= mgcos = 20n, the inertial action of the object continues to travel upward to the velocity is zero. Let this time be t2 and the acceleration is'，a2=8m/s²，t2=，s2=1m。Therefore, the object moves downward from rest after continuing to move upward.

m*a3=f'-f,a3=4m s,1s after v=a3*1=4m s,s3=2m.

3. It refers to the displacement and distance of the object from the bottom to the description of problem (2). s=s1+s2-s3=3m, distance l=s1+s2+s3=7m

1.If you analyze the force on a small wooden block, you don't need to draw other forces after drawing the gravity and support forces. If you are analyzing the car, you need to draw the gravity, as well as the pressure and support force of the wooden blocks on the car.

If you want to analyze the small wooden block and the trolley as a whole, you only need to draw the gravity and support force.

2.Right. 3.Due to the analysis of the small wooden blocks, the car supported him upward. Because the analysis is made of small wooden blocks at this time, it is necessary to draw on small wooden blocks instead of drawing ** segments.

1.The support force must be drawn, otherwise the force will be unbalanced and acceleration will be generated. Pressure is the force of gravity, no need to draw. 2.Think so of it. 3.The force should be drawn on the point of application.

1.A thin tablecloth is laid on the smooth top, and two cups of water are placed on the tablecloth. When the tablecloth is jerked away from the tabletop, the cups on the tablecloth do not move (optionally "would" or "won't"), indicating that the cups are inertial.

2.When Wang Haifeng traveled by long-distance bus, he paid attention to the milestones next to the straight highway and wrote down the corresponding time.

Milestones: 10km, 20km, 30km, 40km, 50km

Observation time: 6:30, 6"45, 7:00, 7:20, 7:39

The average speed of the car in the first 30km is 10 meters per second, 30000 m 50*60s

Question 1: No, there is inertia. Question 2, 30000 (50*60)=10m s

There is friction, and the friction force of the object is equal to the pulling force of the spring, so the speed is constant.

By letting go when the rope is just straightened, the ball B swings downwards from a stationary state, and A does not leave the ground. Therefore b does a circular motion.

Set the ob length L. The velocity v of the ball B by the kinetic energy theorem satisfies mbv 2 = mbglsin30

For ball b from Newton's second law, t-mbgsin30 = mbv l (t is the tension, the force in the longline direction provides the centripetal force).

Pair ball a t-mag=0

The ratio of the mass of ball A to the mass of ball B is ma:mb=3:2

When the angle between the B section of the string and the vertical direction is 60°, the gravity of A provides the centripetal force of B, and the direction of F = MA*G

F direction = vb squared mb r

VB can be calculated by the conservation of mechanical energy

List the equations to find the answer.

Gravity work is combined with circular motion.

The answer is b, because the force in the horizontal direction and the vertical direction does not change (unless the gravity of the trolley changes).

The friction on the ground when the trolley is stationary is 0 because he is in a horizontal position and has no tendency to move.

b Regardless of the stationary motion, the equilibrium force is applied in the vertical direction, the gravity is equal to the supporting force, the gravity is unchanged, and the pressure is unchanged.

There is no friction when at rest.

The question is correct, and you don't need to look at the answer, you think that "it is impossible to have so much energy to make the block rush to point B, because he has been subjected to friction before, so I don't think it can even reach point A", your point of view is correct under the condition that the object is released from point A at rest, because the energy is conserved, and the object cannot reach point A when it is bounced back. But pinch, you probably didn't see this topic clearly, right? "A small block of mass m slides down from a certain vertical downward initial velocity", see no, there is an initial velocity, so as long as the initial velocity is satisfied, it can reach point B.

Pay attention to the examination questions!

"A small mass of m slides down from a certain vertical downward initial velocity".

It is not a stationary fall, it has an initial velocity, and there is no limit to the size. Make the initial velocity large enough, energy is indispensable.

That's right, there was initial velocity at the beginning.

That's right, at the beginning a had muzzle velocity.

Drawing, the magnitude of gravity is never the same.