Newton s second law of motion, Newton s second law of motion

Example: The locomotive with a mass of 100t starts from the parking lot, after 225m, the speed reaches 54kmh, at this time, the driver turns off the engine, let the locomotive enter the station, and the locomotive travels another 125m before stopping at the station. Let the resistance of the locomotive remain unchanged, and seek the traction force of the locomotive before turning off the engine.

Solution: In the acceleration phase: muzzle velocity v0 = 0, final velocity v1 = 54 km h = 15m s, displacement x1 = 225 m, obtained by v -v0 = 2ax, acceleration a1 = v1 2x1 = 15 (2 225) = m s

From Newton's second law, f lead + f resistance = ma1

10 to the fifth power n

5 10 to the fourth power n, deceleration phase, initial velocity v1 = 15 m s, final velocity v2 = 0 , displacement x2 = 125 m

is obtained by v -v0 = 2ax and acceleration a2 = -v1 2x

15 m²/2×125 m/s²

M s minus indicates that the A2 direction is opposite to the V1 direction).

From Newton's second law, f resistance = ma2

10 to the fifth power (

9 10 to the fourth power n

A negative sign indicates that the f-resistance is opposite to the v1 direction).

Therefore, the traction force of the locomotive f lead = -f resistance + ma1 = 5th power n of 10

The derivation of ft=mv-mv-mv (momentum theorem) is derived from at=v-v-v, Newton's second law, and the mass m is multiplied by both sides of the equal sign at the same time, that is, the momentum theorem ft=mv end - mv beginning. It's so hard, extra points, extra points.

b, the box gives the wooden block friction to the right, and the ground gives the box the magic friction to the left, f-f=ma

<>1. We have expounded Newton's first law of motion, and today we will continue to explain Newton's second law of motion. Newton's second law of motion is commonly formulated that the magnitude of an object's acceleration is proportional to the force, inversely proportional to the mass of the object, and proportional to the number of inverted bridges of the object's mass. The acceleration is in the same direction as the square of the applied force.

2. It can be said that as long as people who have some understanding of physics, they know that the unit of mechanics is Newton, referred to as the ox, and the symbol is an uppercase n. Newton's second law states that the acceleration of an object is directly proportional to the net force experienced by the object and inversely proportional to the mass of the object.

The tension of the light rope of the 2m wooden block, the pressure and friction of the M wooden block, the support force and friction of the ground, these are 5, A is incorrect.

t m+2m+3m)*(m+2m)= so the light rope will not break, b is incorrect.

Therefore the light rope will not break, c correct.

t m+2m)*m=1 3t is not 2 3t d is incorrect.

Law content.

The acceleration of an object is directly proportional to the resultant force f experienced by the object, inversely proportional to the mass of the object, and the direction of acceleration is the same as that of the resultant force. From the point of view of physics, Newton's second law of motion can also be expressed as "the rate of change of momentum of an object with time is proportional to the sum of the external forces subjected to it".

Draw first. Both up and down movements can be regarded as uniform acceleration linear motion with an initial velocity of 0.

Let the coefficient of friction be ; The uplink time is t, and the sliding time is 2t

The component of gravity along the perpendicular slope, i.e., the positive pressure n=mgcos

Friction force f= n= mgcos

The component of gravity along the inclined plane f=mgsin

When sliding up along the bevel:

F = F + F = MGSIN + MG COS

a1=f m = gsin + g cos

s=1/2a1t^2= gμcosθ)t^2...

When sliding down along the slope:

F-fit'=f-f=mgsinθ- mgμcosθ

a2=f-combined'/m=gsinθ- gμcosθ

s=1/2a2(2t)^2=2(gsinθ- gμcosθ)t^2...

Obtained from: g cos ) = 2 (gsin - g cos )

sinθ+ cosθ=4(sinθ- cosθ)

3sinθ=5μcosθ

3sinθ/ 5cosθ= 3tanθ/5

Amitabha!!

This topic mainly examines the force analysis of an inclined object and the uniform accelerating linear motion with an initial velocity of 0 s=1 2at 2

Also, sometimes, the deceleration motion can be reversed as the acceleration motion. For example, when the upslide process of this problem is displaced, it can be reversed as a uniform linear motion with an initial velocity of 0.

*10=630 (n)

2), x-630 = 63 * x = (n) (3), which means that Yang Liwei can withstand the vertical ascent of the spacecraft at a maximum of 7g with an acceleration.

4) The feeling of being overweight.

f=63*, which indicates the limit that Yang Liwei can bear.

It's the same feeling as when the spaceship accelerates up, overweight.

。。。Such a simple question, I am dizzy, any sophomore student will do it.

f-umg=ma1

umg=ma2

1 2 (a1-a2) t square = x

0 synthesize the equation on the tree, substitute the data, and calculate it yourself.

Let the acceleration of the block be a, due to the dynamic friction factor u=. So the trolley acceleration is meters per second squared in 2 seconds, and the block moves s=at 2 2=4a 2=2a the trolley moves s=

The displacement difference is s1=s-s=2a-5=1

So a = 3 meters per second squared.

f=ma+ma-car=4x3+

So f=37n is exactly within 2 seconds, and when f>37n, it is also compliant.

The easiest way to solve this problem is to draw a v,t diagram.

In the vertical direction, there is no motion of the object, let alone acceleration, so the force in the vertical direction is balanced. f*sin(θ)g (1)

In the horizontal direction, the object moves at a uniform speed and the acceleration is zero, so the force in the horizontal direction is also balanced. f*cos(θ)g*μ 2)

2) Divide by (1), cotangent( ) so there is =arc cotangent (

With orthogonal decomposition method: because the object is moving at a uniform speed, the net force is zero.

Horizontal FCOS FN

Vertically fsin +fn g

Get f g (cos + sin ) when the denominator is the largest, f is the smallest. Denominator (1 2) square root *sin( +a).

Where, the angle a satisfies the square root of sina 1 (1 2) and cosa 1 2) square root.

Obviously, when sin(+a) 1, f is minimal, then tana=1 and arctan1 = arctan at 90 degrees