There is a car on a horizontal road with a mass of 500 kg, which is subjected to resistance 0 05 tim

1. Because it is stationary, it is a balanced force, that is, the pulling force is equal to the resistance at this time, and the resistance is 200N, and the resultant force is 0N

2. When moving at a constant speed, the tensile force is equal to the movement resistance f=.

3. Accelerate to the left, and the resultant force is 105n to the left.

The point is that the pulling force is equal to the resistance force when the object is pulled but not pulled, and the resistance force is always equal to the pulling force when the object is pulled but not pulled.

Solution: (1) The trolley is stationary, so it is in equilibrium.

According to the condition of the equilibrium of the two forces, the resistance f1 = f1 = 200n, because the two forces are equal in magnitude and opposite in direction, so the net force is equal to 0n;

Answer: The resistance and net force of the trolley are 200N, on 2) by the sliding friction force f2=, when the trolley moves in a uniform linear line, the trolley is in a state of equilibrium, then the tensile force f2=f2=250N;

Answer: The horizontal tensile force is 250N

3) At this time, the resultant force f = 350n-250n = 100n, because the force is the reason for changing the motion state of the object, then the trolley will accelerate to the left

Answer: The trolley will accelerate to the left, and the net force of the trolley is 100N, and the direction is to the left

(1) Sliding friction: f=, so with 200n horizontal tension to pull the stationary trolley, the trolley is still stationary, so it is in a state of equilibrium, obtained by the condition of two forces balance, at this time the trolley is subjected to the resistance f = f=200n;

2) Because the sliding friction force on the trolley is 250N, when the trolley moves in a uniform linear line, the trolley is in equilibrium, then the tensile force f = f = 250N;

3) Because the pressure of the trolley on the ground and the roughness of the contact surface remain unchanged, the friction between the trolley and the ground remains unchanged, and the size is still 250N; And because the pulling force of 350N is greater than the friction force of 250N, the car accelerates the movement

Answer: (1) The resistance of the trolley is 200N (2) The horizontal tensile force is 250N (3) The friction force of the trolley remains unchanged, and the trolley will do an acceleration motion

By the meaning of the title knows that the trolley is inExerciseThis resistance is the sliding friction force, which can be solved by the formula f= fn (fn is the support force) f=250 (g=10m s2).

If the maximum static friction = sliding friction, then when less than 250n force is applied to the trolley, the trolley will not move; At 250N, this is a turning point, i.e., "just in motion, just in motion"; When it is greater than 250N, the trolley moves.

Untie; f=g*

1) When the trolley is stationary, what is the resistance of the trolley when the trolley is pulled by 200N horizontal tension?

The pulling force is less than the resistance required for the movement, and the car remains stationary and is subject to the equilibrium force. So the resistance at this time is also 200n

2) When the horizontal tension is great, can the car do a uniform linear motion?

When the tensile force is balanced with the resistance, the trolley can do a uniform linear motion, so the tensile force should be: f=f=250n

3) When the horizontal tension of the trolley reaches 350N, does the friction of the trolley change? - Unchanged.

What kind of movement do you think the car does at this time? ——The trolley accelerates in a straight line.

(1) When the trolley is stationary, the resistance of the trolley in the horizontal direction is the same as the tensile force, which is a pair of balanced forces, so the magnitude of the resistance is 200N;

2) From this question, in the process of moving on the horizontal road surface, the resistance is twice the weight of the vehicle, so when the car can do uniform linear motion, the tensile force on the car:

f=f=A: (1) When the trolley is stationary, pull the trolley with 200N horizontal tension, and the resistance of the trolley is 200N;

2) When the horizontal tensile force is 250N, the trolley can do uniform linear motion

1. 500kg*g*>>200n,∴f=200n2.According to 1 and Newton's law, it can be seen that the horizontal tensile force required for uniform motion is 245n3

No change, 350N; 350N>245N, so the trolley accelerates evenly.

Halo, forgetting about the drag coefficient.

maximum friction fmax=;

1. Because the tensile force is less than the maximum frictional force, f=f pull=200N;

2. When the horizontal tensile force = maximum friction force fmax = 250N, the trolley can do uniform linear motion;

3. When the trolley is subjected to a horizontal tensile force of 350N, the friction force on the trolley is the maximum friction force, that is, Fmax=250N; At this time, the trolley does acceleration motion a=(f-fmax) m=

The resistance experienced by the trolley is 200N

250N does not change the linear motion with uniform acceleration.

nn3, unchanged. Evenly accelerated movement. Got it?

1.200n, of course.

2.It's 250n hours.

3.At this time, the friction force is 250N, and the trolley accelerates uniformly and moves in a straight line.

(1) 200n (2) 250n (3) 250n accelerated motion test question analysis: (1) sliding friction: f=, so with 200n horizontal tension to pull the stationary trolley, the trolley is still stationary, so it is in a state of equilibrium, obtained by the condition of two forces balance:

At this time, the resistance of the trolley f = f = 200n;

2) When the trolley moves in a uniform linear line, the trolley is in a state of equilibrium, because the sliding friction force on the trolley is 250N, then the tensile force f = f = 250N;

3) Because the pressure of the trolley on the ground and the roughness of the contact surface remain unchanged, the friction between the trolley and the ground remains unchanged, and the size is still 250N; And because the pulling force of 350N is greater than the friction force of 250N, the car accelerates the movement

Constant speed, traction f=drag=

fThe work done in 5 seconds.

w = fs = fvt = 500 * 20 * 5 = 50000 J power p = w t = 10000 watts = 10 kilowatts.

The car moves horizontally, there is no displacement in the direction of gravity vertically downward, and the work done by gravity = 0

Solution: 1. The car is dragged by traction and the drag is 500kg*10n kg*1 10=500n

W=fs=500N*20M S*5S=50000JP=W T=FSt=FV=500N*20M S=10000W2, because there is no distance in the reverse gravity, so the work done by the gravity of the car is 0j

First, the work done by traction is mg*s*20m=50000j

The power is w= f * v=10000w

Second, gravity does not do work. There is no displacement in the direction of gravity.

Resistance = 500*10*

If the horizontal tension of 200N is less than 250N, the trolley will be stationary.

Resistance = Pulling force = 200n (net force is 0).

If the trolley can move, then the resistance =

It can be seen that the trolley does not move.

Received static friction 200 N.

(1) When the trolley is stationary, the resistance of the trolley in the horizontal direction is the same as the tensile force, which is a pair of balanced forces, so the magnitude of the resistance is 200N;

2) From this question, in the process of moving on the horizontal road surface, the resistance is twice the weight of the vehicle, so when the car can do uniform linear motion, the tensile force on the car:

f=f=A: (1) When the trolley is stationary, pull the trolley with 200N horizontal tension, and the resistance of the trolley is 200N;

2) When the horizontal tensile force is 250N, the trolley can do uniform linear motion

f = f-f

The 100n direction is the same as the tensile force.

From the meaning of the title: horizontal struggling sail force f=

The resistance f=f f, so the resultant force slows down the hail f'=f-f= direction, along the horizontal tensile direction, the stove.