An elevator can transport goods with a mass of 1 2 t from the ground floor at a constant speed withi

The elevator transports the goods of mass t from the ground floor to the roof of the fourth floor at a constant speed, and the elevator does work w=gh=j, which is determined by the formula of power. p=wt

w；Answer: The power of the elevator is w

Answer]: B elevator does not have a strict speed classification, and it is customary in China to classify it according to the following methods:

1) Low-speed ladder (Class C ladder). Elevators with a speed lower than are usually used in buildings below 10 floors or passenger and freight elevators or freight elevators.

2) Fast ladder (class B ladder). Elevators with a speed of are usually used in buildings with more than 10 floors.

3) High-speed ladder (Class A ladder). Elevators with a speed greater than , usually used in buildings with a height of 16 storeys or more.

4) Ultra-high-speed ladder. Elevators with a speed of more than are usually used in super high-rise buildings.

Answer]: B2021 textbook p137

Knowledge points: solid Nayou attack conveying equipment and elevators. There is no strict speed classification of elevators, and it is customary to classify them in China according to the following methods:

1) Low-speed ladder (Class C ladder). Elevators with lower speeds are usually used in buildings with less than 10 floors or passenger and freight elevators.

2) Medium-speed ladder (Class B ladder). Elevators with a speed of are usually used in buildings with more than 10 floors.

3) High-speed ladder (Class A ladder cave brother). Elevators with greater speed are usually used in high-rise residential buildings, office buildings, hotels and other buildings with more than 16 floors.

4) Ultra-high-speed ladder. Elevators with excessive speed are usually used in super high-rise buildings.

Very simply, in the first 2s, the elevator does a uniform acceleration linear motion, a = 3, the external force on the object is ma, ma = the supporting force of the elevator on the object - mg, the supporting force is calculated, and then the supporting force of the elevator on the object = the pressure of the object on the floor (note that the two forces are equal in magnitude and opposite in direction).

After 4-7s, the elevator does a uniform deceleration linear motion, a = -2, the combined external force of the object is ma, ma = the supporting force of the elevator to the object - mg, the supporting force is calculated, and then the supporting force of the elevator to the object = the pressure of the object on the floor (note that the two forces are equal in magnitude and opposite in direction).

Is the direction of attention ok?

a is the acceleration, which is the slope of the diagonal line.

The displacement should be the area of the graph. It's very good to ask for it, forget it yourself.

The first 2 seconds of acceleration a=6 2=3

f=m(g+a)

After 4 seconds, acceleration a=

f=m(g+a)

The displacement is divided into three sections.

Just use the formula s=v0*t+

First of all, it is clear that the unit T, the ground floor is the 1st floor, which rises to the 4th floor;

p*t=mgh

10*p=p=12348 (w)=

Known: mass m=

kg, time t=10s, height h=6 3m=18m Find: (1) The work done by the elevator motor to overcome gravity w=?; 2) The power of the elevator motor p=?

1) The gravity of the elevator with the passengers g=mg=

kg× n/kg=

n, the work done by the elevator motor to overcome gravity: w=gh=

n×18m=

j；(2) The power of the elevator motor: p=wt

J10SW=A: (1) The work done by the elevator motor to overcome gravity.

j；(2) The power of the elevator motor is, at least.

When rising and falling at rest and at a constant speed, the force is balanced, and the supporting force = gravity.

The elevator that transports goods carries a weight of 4 x 10 to the power of five n. When the elevator is stationary, the support force of the elevator to the heavy object is (4*10 to the fifth power) n; If the weight is lifted at a constant speed of 1m s, the support force of the elevator for the heavy object is (equal to) the fifth power n of 4*10; If the weight is lifted at a constant speed of 5ms, the support force of the elevator to the heavy object (equal to the fifth power n) of 4*10, and if it is reduced to the first floor at a constant speed of 2ms, the support force of the elevator to the heavy object (equal to the fifth power n.) of 4*10

Solution: t=3min=180s

The total weight of the freight elevator box is: g = mg = 1200 10 = 12000 (n) The increased gravitational potential energy is: w = gh = 12000 54 = 648000 (j) The draw power of the pulling force:

p=w t=648000 180=3600(w) A: Omitted.

I hope it helps you, and if you have any questions, you can ask them

I wish you progress in your studies and go to the next level! (*