Question

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?

Answer 1

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.

Step-by-step explanation:

The acceleration can be calculated by the formula

`a=(F)/(m)`

The force can be calculated by the formula

`F=(kq1q2)/(r^2)`

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

`k=\text{ 9}*10^9\text{ N m}^2\text{ /C}^2`

On substituting the values, the force will be

`\begin{gathered} F=(9*10^9*(96.852*10^(-6))^2)/((3.88)^2) \\ =5.61\text{ N} \end{gathered}`

The acceleration will be

`\begin{gathered} a=(5.61)/(3.24) \\ =1.73\text{ m/s}^2 \end{gathered}`

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