Physics, 12.11.2019 23:31 orlando19882000

Two identical 0.200kg mass are pressed against opposite ends of a light spring of force constant 1.75n/cm compressing the spring by 37 cm from it’s normal length . find the speed of each mass when it has moved free of the spring on a frictionless horizontal table

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This type of a problem can be solved by considering energy transformations. Initially, the spring is compressed, thus having stored something called an elastic potential energy. This energy is proportional to the square of the spring displacement d from its normal (neutral position) and the spring constant k:

$E_p=\frac{1}{2}kd^2= \frac{1}{2}175\frac{N}{m}\cdot 0.37^2m^2=11.98J$

So, this spring is storing almost 12 Joules of potential energy. This energy is ready to be transformed into the kinetic energy when the masses are released. There are two 0.2kg masses that will be moving away from each other, their total kinetic energy after the release equaling the elastic energy prior to the release (no losses, since there is no friction to be reckoned with).

The kinetic energy of a mass m moving with a velocity v is given by:

$E_k = \frac{1}{2}mv^2$

And we know that the energies are conserved, so the two kinetic energies will equal the elastic potential one:

E_p = 2E_k=mv^2

From this we can determine the speed of the mass:

$E_p =mv^2\implies v=\pm \sqrt{\frac{E_p}{m}}=\pm\sqrt{\frac{11.98J}{0.2kg}}=\pm 7.74\frac{m}{s}$

The speed will be 7.74m/s in in one direction (+), and same magnitude in the opposite direction (-).

Answer from: Quest

Answerthe force between the charges will be f/9 (one-ninth of force f)explanationcoulomb's law relates the magnitudes two point charges q₁ and q₂, the distance, r, between the charges, and the electrostatic force, f, between the two charges: $f = \dfrac{k q_1 q_2}{r^2}$ k represents coulomb's constant, k = 8.99×10⁹ n·m²/c².note that if we have r = 3r, then $\begin{aligned} f & = \dfrac{k q_1 q_2}{r^2} \\ f_{3r} & = \dfrac{k q_1 q_2}{(3r)^2} \\ & = \dfrac{k q_1 q_2}{3^2 \cdot r^2} \\ & = \frac{1}{9} \cdot \boxed{\dfrac{k q_1 q_2}{r^2} } \\ & = \frac{1}{9} f & & \left(\text{\footnotesize since $ f = \tfrac{k q_1 q_2}{r^2}$}\right) \end{aligned}$

Answer from: Quest

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.

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