Question

A body of mass \(M\), carrying charge \(Q\), falls from rest from a height \(h\) (above the ground) near the surface of the Earth, where the gravitational acceleration is \(g\) and there is an electric field with a constant component \(E\) in the vertical direction. a) Find an expression for the speed, \(v\), of the body when it reaches the ground, in terms of \(M, Q, h, g,\) and \(E\). b) The expression from part (a) is not meaningful for certain values of \(M\), \(g, Q\) and \(E\). Explain what happens in such cases.

Short answer

Answer: The expression for the speed of the body when it reaches the ground is v_f = \sqrt{2(Mgh + QhE) / M}, where M is the mass, Q is the charge, h is the height, g is the gravitational acceleration, and E is the electric field. This expression is not meaningful when either the mass M <= 0 (not a real mass), or if (Mg + QE) < 0. The latter condition represents a situation where the electric force is so strong (and in the opposite direction) compared to the gravitational force that it prevents the body from falling towards the ground, and thus, there is no "speed when it reaches the ground" to be calculated.

Step-by-step solution

Identify the forces acting on the body

The body is under the influence of two forces: gravitational force and electric force. The gravitational force acting on the body is F_gravity = Mg (downwards), and the electric force acting on the body is F_electric = QE (upwards or downwards, depending on the charges' sign). Since these forces are constant, mechanical energy is conserved.

Set up the conservation of energy equation

As the body falls under the influence of these forces, its potential energy is converted into kinetic energy. We can write the conservation of energy equation as follows: Initial potential energy (gravitational) + Initial potential energy (electric) = Final kinetic energy + Final potential energy (gravitational) + Final potential energy (electric) We will use subscripts i and f to represent initial and final states.

Write expressions for potential and kinetic energies

We have the following expressions for potential and kinetic energies: Gravitational potential energy: U_gravity = Mgy Electric potential energy: U_electric = QEy Kinetic energy: K = (1/2)Mv^2 Note that at the starting point y=h and v=0, while at the point when the body reaches the ground y=0 and we want to find v. Substituting these into the equation from Step 2, we get: Mgh + QEh = (1/2)Mv_f^2

Solve for the final velocity v_f

Solving for v_f, we get: v_f = \sqrt{2(Mgh + QhE) / M} This is the expression for the speed of the body when it reaches the ground in terms of M, Q, h, g, and E.

Analyze cases where the expression is not meaningful

The expression for v_f is not meaningful when the denominator is zero or negative, as the square root of a negative number is not a real number. This indicates a situation where either the mass M, gravitational acceleration g, or the electric field E takes on an unusual value. More specifically, the expression is not meaningful when M <= 0 (not a real mass), or if (Mg + QE) < 0. The latter condition implies that the electric force is so strong (and in the opposite direction) compared to the gravitational force that it prevents the body from falling towards the ground. In this case, the body may stay at the initial height or even move upwards, so there is no "speed when it reaches the ground" to be calculated.

Key concepts

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy possessed by an object due to its position in a gravitational field. For an object of mass M at a height h above the Earth's surface, the GPE is given by the formula:

\( U_{gravity} = Mgh \)
where g is the acceleration due to gravity. GPE is a form of potential energy, the energy stored and ready to be converted into other forms, such as kinetic energy, as the object begins to move. In our exercise, when the body starts at height h, its initial GPE is \( Mgh \), and as it falls towards the Earth, this GPE is transformed into kinetic energy.

Electric Potential Energy

Electric potential energy is the energy a charged object possesses by virtue of its position in an electric field. The formula for the electric potential energy, U_electric, of a charge Q in an electric field E is:

\( U_{electric} = QEy \)
In this case, y represents the position of the charge in the direction of the electric field. As the body with charge Q falls through the constant vertical electric field, its electric potential energy changes. Based on the sign of the charge and the direction of the field, this energy may increase or decrease, influencing the body's total energy as it moves.

Kinetic Energy Conservation

Kinetic energy conservation is a principle based on the conservation of mechanical energy, stating that in the absence of non-conservative forces like friction or air resistance, kinetic energy is conserved. The kinetic energy, K, of an object with mass M and velocity v is given by:

\( K = \frac{1}{2}Mv^2 \)
At the beginning of our example, the body is at rest, so its initial kinetic energy is zero. As it falls, the GPE and electric potential energy convert into kinetic energy. By the time the body reaches the ground, its velocity v represents all kinetic energy it has gained.

Mechanical Energy Conservation

Mechanical energy conservation is a fundamental principle stating that the total mechanical energy (including kinetic energy and all forms of potential energy) in a system remains constant if only conservative forces are acting. In our scenario, the forces involved are gravity and the electric force, both conservative. Therefore, the initial mechanical energy, the sum of gravitational potential energy and electric potential energy, must equal the final mechanical energy, which includes the kinetic energy as the body hits the ground and the reduced potential energy at that point. If additional forces act on the body, causing energy to be lost, the mechanical energy will not be conserved, but this is not the case in our theoretical exercise.