# A 0.40-kg cart with charge 2.5\times 10-5 C starts at rest on a horizontal friction-less surface...

## Question:

A {eq}0.40-kg {/eq} cart with charge {eq}2.5\times 10-5 C {/eq} starts at rest on a horizontal friction-less surface {eq}0.4 m {/eq} from a fixed object with charge {eq}+2.0\times 10-4 C {/eq}. When the cart is released, it moves away from the fixed object.

A) How fast is the cart moving when it is very far (infinity) from the fixed charge?

B) How fast is the cart moving when it is {eq}2.0\ m {/eq} from the fixed object?

## Conservation of Energy:

The Conservation of Energy Principle states that in the absence of external force acting on a system, the total energy of a system will remain constant. For this problem, the electric potential energy due on the charged particles will be equal to the kinetic energy of the particle that moved.

Given:

{eq}m_c = 0.40 \ kg {/eq} Mass of the cart

{eq}q_c = 2.5 \times 10^{-5} \ C {/eq} Charge of the cart

{eq}d = 0.4 \ m {/eq} Distance between the cart and the fixed object

{eq}q_o = 2.0 \times 10^{-4} \ C {/eq} Charge of the fixed object

{eq}k = 9.0 \times 10^{9} \frac {Nm^2}{C^2} {/eq}

Part A) Since both the cart and the fixed object are both stationary initially, then the kinetic energy of the cart-fixed object system is zero. This means that the total energy of the system initially is due to the electric potential energy.

{eq}PE_I = \frac {k q_c q_o}{d} = \frac {9.0 \times 10^{9} \frac {Nm^2}{C^2} * 2.5 \times 10^{-5} \ C * 2.0 \times 10^{-4} \ C}{0.4 \ m} = 112.5 \ J {/eq}

When the cart moved at infinite distance from the fixed object,

{eq}PE_F = \frac {9.0 \times 10^{9} \frac {Nm^2}{C^2} * 2.5 \times 10^{-5} \ C * 2.0 \times 10^{-4} \ C}{\infty} = 0 \ J {/eq}

Using conservation of energy,

{eq}PE_I = PE_F + KE {/eq}

{eq}112.5 \ J = 0 + \frac {1}{2} m_c v^2 {/eq}

{eq}v = \sqrt {\frac {2 *112.5 \ J}{0.40 \ kg}} = 23.71708245 = \boxed {24 \frac {m}{s}} {/eq}

Part B) Using the same principle we used in Part A,

{eq}PE_I = PE_F + KE {/eq}

{eq}112.5 \ J = \frac {9.0 \times 10^{9} \frac {Nm^2}{C^2} * 2.5 \times 10^{-5} \ C * 2.0 \times 10^{-4} \ C}{2 \ m} + \frac {1}{2} (0.40 \ kg) v^2 {/eq}

{eq}112.5 \ J = 22.5 \ J + \frac {1}{2} (0.40) v^2 {/eq}

{eq}v = \sqrt {\frac {2(112.5 \ J - 22.5 \ J)}{0.40}} = 21.21320344 = \boxed {21 \frac {m}{s}} {/eq}