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Mars2501 [29]
2 years ago
8

An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass

1.2 kg is attached to the free end of the spring. The glider is pulled toward the right along a frictionless air track, and then released. Now the glider is moving in simple harmonic motion with amplitude 0.045 m. The motion is horizontal (one-dimensional).
Suddenly, Slimer holding an apple flies in and approaches the glider. Slimer drops the apple vertically onto the glider from a very small height. The apple sticks to the glider. The mass of the apple is 0.18 kg.
Recall that the total mechanical energy is E = 1/2 mv^2 + 1/2 kx^2 = 1/2 kA^2 = constant
(a) Calculate the new amplitude of the motion of the glider with apple if the apple is dropped at the moment when the glider passes through its equilibrium position, x = 0 m.
Hints: The total energy of the glider just before the collision is E = 1/2mglider v^2 = 1/2KA^2
The apple sticks to the glider in a completely inelastic collision. The glider is now moving with the apple but at a lower speed. The linear momentum is conserved. Write a corresponding equation.
Also, assume that the collision is very short, so just before the collision the glider is at x = 0 m, and just after the collision the glider and apple are still at x = 0 m. Therefore, the total energy of the glider just after the collision is Enew = 1/2mglider + applev^2new = 1/2kA^2new
(b) Calculate the period of the motion of the glider and the period of the motion of the glider with apple.
Hint: it's a very simple question.

Physics
1 answer:
solniwko [45]2 years ago
4 0

Answer:

A) The new amplitude = 0.048 m

B) Period T = 0.6 seconds

Explanation: Please find the attached files for the solution

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1. Determina el momento que produce una fuerza de 7 N tangente a una rueda de un metro de diámetro, sabiendo que el punto de apl
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Answer:

τ= F r     into the blade

Explanation:

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         τ = F x r

where the bold indicates vectors

Let us write in the expression in magnitude

         τ = F r sin θ

in our case the force is tangent to the wheel therefore the angle between F and the radius is 90º, and the sin 90 = 1

       τ= F r

The direction of τ can be used by the rule of the right hand, the fingers curve in the direction of the torque when advancing from the force to the radius and the thumb points in the direction of the torque.

In this case, for a clockwise rotation, the fingers are curved in the direction and the thumb points into the blade, this is the direction of the τ.

TRASLATE

El momento de una fura es definido por

         τ = F x r

donde la negrillas indican vectores

Escribamos en ta expresión en magnitud

          τ = F r sin θ

en nuestro caso la fuerza es tangente a la rueda por lo tanto el angulo entre F y el radios es 90º, y el sin 90=1

        τ = F r

la dirección de tau la podemos  usar la regla de la mano derecha, los dedos curva en la dirección del torque al avanzar dese la fuerza al radio y el pulgar apunta en la dirección del torque.

En este caso para un giro en sentido horario los dedos se curvan ente sentido y el pulgar apunta hacia dentro de lla hoja, esta es la dirección del troque

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