Question

Two 2.0 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge. PART A

What is the magnitude of the electric force on one of the masses?
Express your answer to two significant figures and include the appropriate units.

PART B
What is the initial acceleration of the mass if it is released and allowed to move?
Express your answer to two significant figures and include the appropriate units.

Answer 1

The magnitude of the electric force on one of the masses is 102.71 N. The initial acceleration of the mass is 51.36 m/s^2.

Step-by-step explanation:

PART A:

To find the magnitude of the electric force on one of the masses, we can use Coulomb's Law.

The formula for the magnitude of the electric force is:

F = k * (|q1| * |q2|) / r^2

where F is the force, k is the Coulomb's constant (9 x 10^9 Nm^2/C^2), |q1| and |q2| are the magnitudes of the charges (9.6 μC), and r is the distance between the charges (1.1 m).

Plugging in the values:

F = (9 x 10^9 Nm^2/C^2) * (9.6 μC * 9.6 μC) / (1.1 m)^2

F = 102.71 N

The magnitude of the electric force on one of the masses is 102.71 N.

PART B:

To find the initial acceleration of the mass when it is released and allowed to move, we can use Newton's second law.

The formula for the acceleration is:

a = F / m

where a is the acceleration, F is the force (102.71 N), and m is the mass (2.0 kg).

Plugging in the values:

a = 102.71 N / 2.0 kg

a = 51.36 m/s^2

Answer 2

A) Force = 0.69 N

B) Acceleration = 0.34 m/s^2

Step-by-step explanation:

The electric force is given by the expression:

`F_e= K *(q_1*q_2)/(r^2)`

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q1 and 12 is the charge of the particles, and r is the distance:

`F_e = 9*10^9 Nm^2/C^2 * ((9.6*10^(-6)C)^2)/((1.1m)^2) = 0.69 N`

Part B.

For the acceleration, you need Newton's second Law:

F = m*a

Then,

`a = (F)/(m) = (0.69 N)/(2 kg) = 0.34 m/s^2`