The rated power of a crane is 10kW. It is used to lift a 2.5 × 10 ^ 3kg steel beam. At first, the steel beam moves up uniformly with an acceleration of 0.2m/s? 2 for 1s, and then it moves at a constant speed after adjustment (1) The instantaneous power of pulling force at the end of 1s; (2) What is the maximum speed that the steel beam can reach when it rises? (g = 9.8 m / S 2)

The rated power of a crane is 10kW. It is used to lift a 2.5 × 10 ^ 3kg steel beam. At first, the steel beam moves up uniformly with an acceleration of 0.2m/s? 2 for 1s, and then it moves at a constant speed after adjustment (1) The instantaneous power of pulling force at the end of 1s; (2) What is the maximum speed that the steel beam can reach when it rises? (g = 9.8 m / S 2)

(1) If the tension of the crane is f and the beam weight is g, then a = (F-G) / m, then f = m * a + G = 25000n, instantaneous power P = f * V, v = a * t = 0.2m/s can be obtained, and P = 5000W = 5kW can be obtained
(2) According to the meaning of the question, if the final motion is uniform, then f = g = 24500n, and rated power P = 10kW, from P = f * V, v = P / F = 0.4m/s

Physical friction exercises for senior one A body with mass of 5 kg is pressed on a vertical wall with horizontal force F = 20n. The object slides down along the wall. The dynamic friction coefficient between the object and the wall is μ = 0.3. The direction of the friction force on the object is judged and the magnitude of the friction force is calculated?

According to Newton's third, the body is supported by a force of 20n, f = μ f = 0.3 × 20 = 6N
Because the slide V goes down, the friction goes up
(in the vertical direction, the object is subject to friction and gravity, and the resultant force F vertical is generally downward, f vertical = 50-6 = 44N)

When people get on a bicycle and move forward, the front and rear wheels of the car are subject to the friction force from the ground A is all forward, B is backward C front wheel forward, rear wheel backward C front wheel backward, rear wheel forward

The friction force is always opposite to the direction of the relative motion of the object
The rear wheel is the driving wheel. Through the chain, the part of the rear wheel contacting the ground tends to move backward, so the friction force is forward, which is the reason why the bicycle will move forward
The front wheel is a driven wheel, which was originally stationary, but the bicycle tends to move forward, that is, the part of the front wheel contacting the ground tends to move forward, so the friction force of the front wheel is backward

A physics problem about friction The car with mass m = 8kg stops on a smooth horizontal surface. A right horizontal constant force of F = 8N is applied to the right end of the car. When the speed of the car to the right reaches 3m / s, a small object with M = 2kg (initial speed is zero) is gently placed on the right end of the car (initial speed is zero). The dynamic friction coefficient between the object and the car is 0.2. Assuming that the car is long enough (g = 10m / S2) (1) What is the acceleration of the small object and the car before it is stationary relative to the car after it is put on the car? (2) How long does it take for a small object to be relatively stationary with the car?

At the beginning: the car is subjected to a force of 8N to the right, so the acceleration is 1m / S2 (to the right)
Later: an object was put on the car, and the initial speed of the object was 0, so the object moved to the left relative to the car, and was subjected to a right friction force of the car on it. The force was n * μ = 20 * 0.2 = 4N
The acceleration of the object is 4N / 2kg = 2m / S2
Then, according to the force and reaction force, the car is given a force of 4N to the left by the object. Therefore, the resultant force of the car is 4N. If the direction is right, the acceleration of the car is
4N/8kg=0.5m/s2
(2) Let the time be t,
When the object and the car are relatively stationary, their accelerations are the same
So 0 + 2T = 3 + 0.5T
T = 2
I'm just a freshman in high school. I want to be right,

Is a glass bottle subject to friction under the following conditions? If there is friction, what is the direction of friction? (1) The bottle is still on the table (2) The bottle was still on the tilted table (3) The bottle is held in the hand with the mouth up (4) There is a piece of paper on the bottle. Hold the bottle and pull it out

(1) The bottle is still on the rough horizontal surface. There is no movement and movement trend of the bottle relative to the horizontal plane, so there is no friction force
(2) The bottle is still on the inclined table. Due to the gravity of the bottle, the bottle tends to move downward along the inclined plane, so the bottle is subject to static friction force upward along the inclined plane
(3) When the bottle is held in the hand and the bottle mouth is downward, due to the gravity of the bottle, the static friction force is upward
(4) The bottle presses a piece of paper, blocks the bottle and pulls out the paper. The bottle and the paper have elastic force and relative motion, and the contact surface is rough. Therefore, there is sliding friction force, and the direction is the same as that of the paper strip

The object a with a mass of 2 kg is stacked on a 4 kg long board. The dynamic friction coefficient of a and B is 0.5. Under the action of horizontal plane tension F, it starts to move from rest on the smooth plane. When the tension f is 12n, the friction heat between AB and ab is? When the tension is 18N, the friction heat is? (the picture shows that the wood block a is placed on the long board B and the tension f acts on a)

When the limit is 15N 12n, the static friction is 18N, and the sliding friction is

1. The rated power of a tractor is 1.5x10 ^ 4W, and the resistance in motion is 1.0x10 ^ 3N. What is the maximum speed of the tractor? If it runs at a constant speed of 5m / s, what is the actual power consumed by the engine? 2. With constant power P = 10kW, the crane lifts the object with mass of 500kg on the ground from static to upward, H = 2m, and reaches the maximum speed

1. Maximum driving speed = power ＋ resistance = 15m / s
Actual power consumption = speed × resistance = 5 × 10 ^ 3W
2. Maximum speed = power △ gravity = 2m / S
Output power × time = added energy of object = gravitational potential energy + kinetic energy
= 500 × 10 × 2 + 500 × 2 △ 2 = 11000j
So time = 11000 △ 10000 = 1.1s

When a body with mass m starts to move in a straight line from rest under the action of a constant combined external force F, then the instantaneous power of force F at time t is p =?, P1 of F in time t =? What is the work done by F force in time t?

If the acceleration is a and the velocity is v when t, then v = at = ft / m, then the instantaneous power of F force at t time p = FV = f ^ 2T / m, and the moving distance S = VT / 2, then P1 = FS / T = f ^ 2T / (2m) of F in time t, w = FS = f ^ 2T ^ 2 / (2m) of F force in time t

In the derivation formula P = FV, when does f use traction force and when does it use resultant force~ Also, by the way, good people mentioned what should be paid attention to in this section~

Traction is the total work, P total
P is work, P is useful
Friction work is p total minus P is useful

Under constant rated power, the train with mass of 300000kg starts from stationary along the straight track and has constant resistance during operation. After 1000s, the maximum running speed reaches 72km / h. at this time, the driver turns off the engine and the train continues to slide for 4km to stop 1. Train acceleration after engine shutdown; 2. Resistance of train running; 3. Rated power of train;

Work done by train in 1000s w = Pt
Converted into kinetic energy w = 1 / 2mvv
And overcome resistance to do 12km work w = FS have Pt = 1 / 2mvv + FS -------- (1)
P = FV -------- (2)
Bring (2) into (1)
Then substitute M = 300000kg
t=1000s
s=12000m
v=72km/h=20m/s
F = 75000n
P=1500000W