When did you learn the physical conveyor belt problem in the first year of high school?

One of the motion models in the chapter of Newton's laws of motion.

First of all, the object is uniform so it will not be subjected to friction, and then, because O1 is the driving wheel, then the main driving force of the belt movement is the static friction of the driving wheel on the belt, this force is the belt at the P point, and the belt is also moving clockwise, so the static friction of the driving wheel at the P point is downward.

Finally, let's look at the friction force of the Q point, because O2 is driven by the wheel, that is to say, O2 will not actively rotate and O2 can only rotate by the static friction of the belt to it, since O2 rotates clockwise, then at the Q point, the static friction of the belt O2 is upward, and in turn the static friction of O2 to the belt is downward.

This topic is similar to the friction between the front and rear wheels of the bicycle, it is a kind of question, it seems that you still need to practice the friction judgment in junior high school.

When the driving wheel is clockwise, it is subjected to counterclockwise friction, and due to the interaction force, the belt is subjected to clockwise friction. So the p-point belt is subjected to downward friction; The driven wheel is driven by the belt, and at the q point, the driven wheel prevents the belt from moving upwards due to friction, so the belt is subjected to downward friction at the q point. And the object moves with the belt, so there is no tendency of relative motion between the object and the belt, so there is no friction, that is, the friction is 0.

So choose D

D is right, point P is the driving wheel to transmit force to the belt, the force is down, point Q is the belt to the driven wheel, the driven wheel is stressed up, the belt is stressed down.

Hello. Pick B

A is relatively stationary with the conveyor belt, and when the conveyor belt with an inclination angle accelerates and rotates with acceleration A in the counterclockwise direction, A has an acceleration A along the inclined downward surface, and the force analysis of A shows that only A hopes that mine will be helpful to you.

If in doubt, keep asking.

In the process of decreasing the block velocity from v0 to v, mg*sin37 degrees + mg*cos37 degrees m*a1, a1 is the acceleration magnitude.

A1 g*(sin37 degrees + *cos37 degrees) 10*( m sec 2

The time it rises along the belt is set to t1, and v v0 a1*t1 gives 4 6 10*t1, t1 seconds.

The distance to slide up is s1 [(v0 v) 2]*t1 [(6+4) 2]*meters (multiply time by average speed).

When the speed of the block reaches 4 m s, it will move downward relative to the conveyor belt because the friction force is less than the downward component of gravity along the conveyor belt, and the acceleration is set to a2, from mg*sin37 degrees mg*cos37 degrees m*a2

Get a2 g*(sin37 degrees *cos37 degrees) 10*( m seconds 2

With the ground as a reference, the movement behind the block is similar to the "vertical upward throwing motion", and the distance it can then glide upwards is set to s2, and v 2 2*a2*s2 gets 4 2 2*2*s2 ,s2 4 meters, and it can be seen that the block has not slid out of the conveyor belt.

If you consider the entire rest of the process as a stage, let its time be t2, then.

s1 v*t2 , that is, 1 4*t2 , the solution of t2 (2 root number 5) seconds seconds, all the time t total t1 t2 seconds.

(1) The acceleration of the workpiece just placed on the horizontal conveyor belt is a1, which is obtained by Newton's second law mg ma1 (1 minute) and solved to obtain a1 g 5 m s2 (1 minute) The time through t1 is the same as the speed of the conveyor belt, then t1 0 8 s (1 minute) The forward displacement is x1 a1t 1 6 m (1 minute) After that, the workpiece will move with the conveyor belt at a uniform speed to point b, and the time t2 0 6 s (1 minute) So the time taken by the workpiece to reach point b for the first time t t1 t2 1 4 s (1 minute) (2) Let the maximum height of the workpiece rise be h, and the kinetic energy theorem is obtained by ( mgcos mgsin ) 0 mv2 (3 minutes) to obtain h 2 4 m (2 points) (3) The time for the workpiece to move upward along the conveyor belt is t3 2 s (1 minute) After that, because the acceleration of the workpiece is the same when moving in the inclined section of the conveyor belt, and the acceleration when moving in the horizontal section of the conveyor belt is also the same, the workpiece will do reciprocating motion on the conveyor belt, and its period is t t 2t1 2t3 5 6 s (1 minute) The time required for the workpiece to move from the beginning of the movement to the first return to the horizontal part of the conveyor belt and the speed becomes zero t0 2t1 t2 2t3 6 2 s (1 minute) and 23 s t0 3t (1 minute) This means that after 23 s the workpiece moves exactly to the horizontal part of the conveyor belt, and the speed is zero, so the workpiece is on the right side of point A, and the distance to point A x lab x1 2 4 m (1 point).

Let the mass of the chalk head be m, and the friction factor with the conveyor belt is u

When the conveyor belt moves at a constant speed.

The chalk head acceleration is: a=umg m=ug

The velocity of the end of the chalk tip is 2m s

Therefore, ug*t=2

Solution: t=4s

ug = when the conveyor belt is evenly decelerated, the chalk head first accelerates to the same speed as the conveyor belt, and then decelerates to 0

The chalk head accelerates evenly and runs backwards relative to the conveyor belt.

Depending on the moment at which the conveyor belt is at the same speed as the chalk head, there is:

ugt1=2-t1

t1=4/3

At this time, the speed of the conveyor belt and the chalk head is: v=2-t1=2 3

At this time, the running distance of the conveyor belt is: s1=(2+v) 2*t1=(2+2 3) 2*(4 3)=16 9

The running distance of the chalk head is: s2=

At this time, the chalk head is running backward from the conveyor belt: s3 = 16 9-4 9 = 4 3

After this, the chalk head velocity is greater than that of the conveyor belt, and the chalk head moves forward relative to the conveyor belt.

The running distance of the conveyor belt is: s4=(2 3+0) 2*t2=(2 3) 2*(2 3)=2 9

The time it takes for the chalk head to slow down to 0 is: t2=v (ug)=(2 3).

The deceleration distance of the chalk head is: s5=(v+0) 2*4 3=(2 3) 2*4 3=4 9

The relative displacement of the chalk head in the deceleration phase and the conveyor belt is: s6=s5-s4=4 9-2 9=2 9

Therefore, the length of the scribing line of the chalk head on the conveyor belt is 4 3m

The whole movement situation is that the chalk head moves 4 3m backward relative to the conveyor belt, and then the chalk head moves 2 9m forward relative to the conveyor belt, and the total stroke is 4 3m

(1) In the process of sliding along the arc track.

mgr=mv 2 2 gives v=3 the acceleration of the motion of the object on the conveyor belt a=gu=2

The maximum distance to the left is s=v 2 2a=

2) The time when the object moves to the left on the conveyor belt t1 = v a = when the object moves to the right at a speed of v, the distance that has moved to the right s1 = v 2 2a = 1 time taken t2 = v a = 1

Time of uniform motion t3=(s-s1) v=t=t1+t2+t3=

v=3m/s

Sliding friction to the right f = a=2m ss3=at

t=(1)2) a=2m ss first accelerates and then has a constant velocity2=at t1=1s

t2=t= 1 +

Ideas for analyzing conveyor belt problems:

Initial conditions Relative motion Judge the magnitude and direction of sliding friction Analyze the magnitude and direction of the resultant external force and acceleration of the object Analyze the change of the future force and motion state by analyzing the relative motion of the object from the change of velocity.

The difficulty is whether the object can remain relatively stationary with the belt when the velocity of the object and the belt is equal in size and direction. Generally, the hypothetical method is adopted, and if it can be established, the key depends on whether the f-static is between 0 and fmax.

Good luck with your studies!

Hope it helps.

And adopt me as.

Thank you!

Yes, the acceleration a=umg m=2, the velocity to zero, the displacement is 1 2at 2 = another method.

From sliding down to a velocity of 0, the frictional force does the work, umg*s=mgr, s=r u=the result is the same.

The second question is to pay attention to the acceleration to 2m s, the time to 1s, the displacement from time is 1m, and the remaining time is, in total.

1. Counterclockwise movement, the bag moves to the right relative to the conveyor belt, so the friction direction is to the left, f=, so a=. The package makes a uniform deceleration motion with an acceleration of 6 to the right. Let b be the velocity v, 10 2-v 2 12=8, so v=2.

2. Clockwise movement, at the beginning, the package is still moving to the right relative to the conveyor belt, and the direction of friction is to the left, as above a=6. The package makes a uniform deceleration motion with an acceleration of 6 to the right. When the bag decelerates to 8m s, the relative motion is gone, the friction disappears, and the bag continues to move in a uniform straight line to the right.

When moving with uniform deceleration: 10-6t=8, t=1 3s. The distance elapsed at this point is:

100-64/12=3m。The remaining 5m does a uniform linear motion. Time 5 8.

The total time is 1 3 + 5 8

The duffel bag moves from left to right under the action of friction. The acceleration is 6 m s2, so it takes 4 3 seconds to increase the velocity to 8 m, and the displacement is less than 8 m, so after 4 3 seconds, it will move in a uniform straight line at a speed of 8 m s.

The rest of the displacements can also be solved by the law of uniform linear motion.

Dude, shouldn't you have a picture of this?

f is the same as the direction of movement of the object, and when v of the object = v of the conveyor belt. f is opposite to the direction of motion, and finally the object is not subjected to force due to inertia.