Analysis of the action of spring constructs in the integrated outer layer

Set at a certain time t, the object is located at x, at this time the combined force it receives is double watt x) yin gsin eve, according to Newton's second law has double x) yin = ma system of gravity potential energy type = Yin g The total potential of the Gesino system is 2, +En, +mgrsin0 uses (l), and the total hall 2), 2, which means that: the balance is the potential position of the potential energy, and the object is at any position. At the time, the potential energy of the system, this potential energy is the total potential energy of the system, which includes the gravitational potential energy and the elastic potential energy, and the total potential energy of the system appears only in a single form.

This total potential energy is called potential energy. Mechanical energy conservation When the spring vibrator moves on a smooth slope, the mechanical energy of the system is conserved. For example, the light springs are used to connect the upper and lower horizontal boards A and B with the masses of respectively = 0.1 kg and m, = 0.2 kg. On the ground, ask for a large downward force F on A, in order to suddenly remove the force, so that A jumps and can just pull up B. No. 2 (3) where A is the amplitude of the spring oscillator. There are two potential energies in equation (3), which are more cumbersome and error-prone, which is not conducive to simplifying the problem. In fact, bringing the formula (2) into the formula (3), you can get the expression of the mechanical energy conservation of the system when the spring vibrator moves on the smooth slope with the Xixin 2 bucket m guard 2 Li 2 2 (4) The mechanical energy conservation of the spring oscillator on a smooth horizontal surface is such that the problem of the spring oscillator on the smooth bevel is simplified.

Please see the following example. Example 1 The light spring is placed on a smooth inclined surface with an inclination of 450. The upper end is fixed, and the lower end is hung with an object of mass 0.2kg, and the spring is extended by 0.lm.

Let the object be a simple harmonic vibration with an amplitude of 0.04m on the inclined surface. When the object moves to the maximum displacement below, it is completely inelastic with an object with a mass of 0.2kg and a speed of vo=08 s. Collision, try to find the amplitude of the system vibration after collision.

The vertical slanted surface of the bullet-like vibrator on the smooth slope of the bullet-like vibrator diagram: when the mass of the object is 0.2kg, the spring is elongated o.lm, we can know that kxo.l=0.2x9.8sin450 has k=13.86 gamma/m) The momentum conservation before and after the object collision is m==Zmv. After the 0.4 (resistance to s) collision, the elongation of the spring vibrator, the two Australian wire 0.2 (m) with the new equilibrium point as the coordinate origin, the relevant data is substituted into the Li Hall.

Bucket solution, 22rZ solution: Set the coordinate system based on the balance position of the board A on the spring, and set the coordinate system of the board A at the equilibrium position to be x0. = The foot call (1) is the initial state when the force F is applied. When the force F is removed, the spring elongation is maximum (A jumps and just pulls up B). The initial state: the initial state: the end state of the Exl field number tree.

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