Two 3.24 kg masses are 3.88 m apart on a frictionless table. Each has 96.852 microCoulombs of charge. What is the initial acceleration of each mass if they are released and allowed to move?

Respuesta :

Given:

The mass is m1 = m2 = 3.24 kg

The distance between them is r = 3.88 m

The charge on each mass is q1 = q2 = 96.852 micro Coulombs

To find the acceleration of the mass.

Explanation:

The acceleration can be calculated by the formula

[tex]a=\frac{F}{m}[/tex]

The force can be calculated by the formula

[tex]F=\frac{kq1q2}{r^2}[/tex]

Here, k is the Coulomb's constant whose value is

[tex]k=\text{ 9}\times10^9\text{ N m}^2\text{ /C}^2[/tex]

On substituting the values, the force will be

[tex]\begin{gathered} F=\frac{9\times10^9\times(96.852\times10^{-6})^2}{(3.88)^2} \\ =5.61\text{ N} \end{gathered}[/tex]

The acceleration will be

[tex]\begin{gathered} a=\frac{5.61}{3.24} \\ =1.73\text{ m/s}^2 \end{gathered}[/tex]

Thus, the acceleration of each mass is 1.73 m/s^2