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"Comprehensive Application of Newton's Second Law of Motion" Newton's Law of Motion PPT
Part One: Learning Objectives
1. Clarify two types of basic problems in dynamics.
2. Master the basic ideas and methods of applying Newton's laws of motion to solve problems.
Comprehensive application of Newton's second law of motion PPT, part 2: basic review
1. Two types of problems in dynamics
(1) Calculate the motion condition given the force conditions: first find a according to Newton’s second law, and then use the kinematic formula to find the motion condition. (such as v0, v, s, t, etc.)
(2) Calculate the force situation given the motion situation: first find a according to the kinematic formula, and then find the force according to Newton's second law.
2. Thought process for dealing with two types of problems
3. Common methods for solving problems
(1) Vector synthesis method: When an object is acted upon by only two forces, apply the parallelogram rule to find the resultant force of the two forces, and then use Newton's second law to find the magnitude and direction of the object's acceleration. The direction of the acceleration is The direction of the resultant external forces on the objects is the same.
(2) Orthogonal decomposition method: When an object is acted upon by multiple forces, the orthogonal decomposition method is often used to find the resultant force of the object, and then Newton's second law is applied to find the acceleration. In practical applications, the force is often decomposed, and the direction of acceleration is selected as the x-axis or y-axis direction. Sometimes the acceleration can also be decomposed, that is, Fx=maxFy=may.
Comprehensive application of Newton's second law of motion PPT, the third part: teacher-student interaction
Known object’s motion due to forces
Key ability 1: Determine motion conditions from forces
(2019•Zhejiang Huzhou High School Grade 1) Skating cart is one of the favorite ice entertainment projects for children. As shown in the picture, Xiao Ming’s mother is playing with Xiao Ming on the ice. The total mass of Xiao Ming and the ice cart is 40 kg. The ice cart The kinetic friction factor between the ice cart and the ice surface is 0.05. In a certain game, assume that Xiao Ming’s mother exerts a horizontal thrust of 40 N on the ice cart. After 10 seconds, the ice cart starts to move from rest, and then stops applying force, causing the ice cart to move. The car slides freely (assuming that the ice car always moves in a straight line during the movement, and Xiao Ming never exerts any force). beg:
(1) The maximum speed of the ice truck;
(2) The total sliding displacement of the ice truck during the entire movement.
[Ideas] (1) From the question, we know that the ice car first makes a uniform acceleration motion and then a uniform deceleration motion. When Xiao Ming’s mother stops exerting force, the speed is maximum, which is obtained by Newton’s second law. Acceleration, use the velocity formula to solve for the maximum velocity.
(2) Calculate the displacement of uniformly accelerated motion according to the displacement formula, remove the force and the ice car performs a uniform deceleration motion process, obtain the acceleration according to Newton's second law, and obtain the sliding displacement according to the kinematic velocity-displacement relationship, which can be obtained total displacement.
Problem solving steps
(1) Determine the research object, conduct force analysis and motion analysis on the research object, and draw a force diagram of the object;
(2) According to the method of synthesis and decomposition of forces, find the total external force (including magnitude and direction) acted on the object;
(3) Set up equations according to Newton’s second law to find the acceleration of the object;
(4) Combined with the given initial conditions of object motion, select the kinematic formula and find the required motion parameters.
Key Capability 2 Isochronous Circle Model
As shown in the figure, ad, bd, and cd are three fixed smooth thin rods in the vertical plane. a, b, c, and d are located on the same circle. Point a is the highest point of the circle, and point d is the lowest point. Each rod is equipped with a small slip ring (not shown in the picture). The three slip rings are released from a, b, and c respectively (the initial speed is 0). Each slip ring is represented in sequence by t1, t2, and t3. The time it takes to reach point d, then ()
A． t1t2>t3
C. t3>t1>t2 D. t1=t2=t3
Key Capability 2 Isochronous Circle Model
As shown in the figure, ad, bd, and cd are three fixed smooth thin rods in the vertical plane. a, b, c, and d are located on the same circle. Point a is the highest point of the circle, and point d is the lowest point. Each rod is equipped with a small slip ring (not shown in the picture). The three slip rings are released from a, b, and c respectively (the initial speed is 0). Each slip ring is represented in sequence by t1, t2, and t3. The time it takes to reach point d, then ()
A． t1t2>t3
C. t3>t1>t2 D. t1=t2=t3
Isochronous circle model
(1) The time it takes for a particle to slide from rest to the lowest point of the ring along different smooth chords on a vertical ring is equal, as shown in Figure A;
(2) The time it takes for the particle to slide from rest to the lower end along different smooth chords from the highest point on the vertical ring is equal, as shown in Figure B;
[Exercise to achieve standards]
1. (2019•Jiangsu Yangzhou High School Grade 1 Mid-term) As shown in the figure, steel components A and B are stacked on the horizontal bed of a truck. The kinetic friction factor between the truck bed and B is μ1, and the kinetic friction factor between A and B is μ2, μ1 >μ2. The maximum acceleration of truck braking is a, a>μ1g. It can be considered that the maximum static friction force is equal to the sliding friction force. When a truck encounters an emergency while driving along a straight road, it is required to stop safely within s0 distance after braking. , the truck’s speed cannot exceed ()
A.2as0 B. 2μ1gs0
C.2μ2gs0 D. (μ1+μ2)gs0
2. (2019·Guizhou Zunyi High School End of the First Period) As shown in the figure, the fixed smooth circular track located in the vertical plane is tangent to the horizontal plane at point M, and is tangent to the vertical wall at point A. The angle between the line connecting points B and M and the horizontal plane is 60°. C is the center of the circular orbit. It is known that at the same time: two balls a and b start from rest at points A and B and follow smooth inclined straight orbits. Move along AM and BM to point M respectively; ball c falls freely from point C to point M. but()
A． Ball a reaches point M first
B． Ball c reaches point M first
C. Ball b reaches point M first
D. Both ball b and ball c may arrive at M first
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