When an arrow is released, where does it obtain its kinetic energy from?

A stretched bow contains potential energy due to its altered shape. When the bow is released to shoot an arrow, this potential energy transforms into the kinetic energy of the arrow.

Understanding Potential Energy: Concepts and Applications

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Potential Energy

It is a key concept in physics that describes the stored energy of an object based on its position or arrangement. The most prevalent form is gravitational potential energy, which is influenced by an object's height relative to a reference point and its mass. Additionally, there are other types, such as elastic potential energy, found in stretched or compressed springs, and chemical potential energy, which is held within the bonds of molecules.

1.0Definition of Potential Energy

It is the energy due to its position or arrangement within a conservative force field.
Potential energy of a body at any position in a conservative force field is defined as the external work done against the action of conservative force in order to shift it from a certain reference point (PE = 0) to the present position.
Potential energy of a body in a conservative force field is equal to the work done by the conservative force in moving the body from its present position to reference position.
At a certain reference position, the potential energy of the body is assumed to be zero.
Potential energy is a relative quantity.
Potential energy is defined only for conservative force fields.

2.0Examples of Potential Energy

A body resting on the roof of a building possesses potential energy. When it is allowed to fall, this energy can be converted into work as it descends.
In a toy car, the wound spring stores potential energy. As the spring unwinds, this potential energy is converted into kinetic energy, propelling the toy car forward.

3.0Relationship between conservative force field and potential energy

$F = - \nabla U = - grad (U) = \frac{\partial U}{\partial x} \hat{i} - \frac{\partial U}{\partial x} \hat{j} - \frac{\partial U}{\partial x} \hat{k}$

If force varies with only one dimension (say along x-axis) then,

$F = - \frac{d U}{d x} = - F d x$

$\Rightarrow \int_{u_{1}}^{u_{2}} d U = - \int_{x_{1}}^{x_{2}} F d x \Rightarrow Δ U = - W_{c}$

Potential energy may be positive or negative or even zero

Potential energy is positive, if the force field is repulsive in nature.
Potential energy is negative, if the force field is attractive in nature.
If r $↑$ (separation between body and force centre), U $↑$ force field is attractive or vice-versa.
If r $↑$ , U $↓$ , force field is repulsive in nature.
IF PE $↑$ , when positive work is done by the external force (against the conservative force)
IF PE $↓$ , when positive work is done by the conservative force.

4.0Gravitational And Spring Potential Energy

Gravitational Potential Energy

when a body is taken against Earth's Gravitational Force then work done by the weight on the body, W_g = –mgh
Earth's Gravitational Force is a conservative force hence this work must be stored in the body in the form of Gravitational Potential Energy $(Δ U_{g})$
From the concept of potential energy

$Δ U = - W_{Conservative} \Rightarrow Δ U_{g} = - W_{g} \Rightarrow U_{f} - U_{i} = - (- m g h)$

$U_{f} - U_{i} = m g h$

$U_{f} \to$ Gravitational Potential Energy at height h

$U_{i} \to$ Gravitational Potential Energy on ground

Here we can assume U_i = 0 for our convenience as potential energy is a relative quantity.
Thus, we have following relation $U_{f} = m g h$

NOTE- For regularly shaped uniform bodies, the potential energy change can be calculated by considering their mass to be centred at the geometrical centre point.

For Example: For a uniform vertical rod of length L, $GPE = m g \frac{L}{2}$

Spring Potential Energy

Work done by spring force $W_{s} = \frac{1}{2} k (x_{i}^{2} - x_{f}^{2})$
As we can see, work done by spring force depends only on initial and final position. Thus, it is also a conservative force and hence Spring Potential Energy (U_s) can be defined as
$U_{s} = - W_{s} = \frac{1}{2} k (x_{f}^{2} - x_{i}^{2})$

5.0Types of Equilibrium

Equilibrium-At equilibrium net force is zero. If net force acting on a body is zero then the body is said to be in equilibrium and positions where the body achieves equilibrium are called equilibrium positions.

At equilibrium, $F = - \frac{d U}{d x} = 0$

Potential Energy Curve-It is a curve which shows the change in potential energy with the position of a particle.

Stable Equilibrium

After a particle is slightly displaced from its equilibrium position, if it tends to come back towards equilibrium due to restoring force then it is said to be in stable equilibrium.
A restoring force is a force that acts to return a body to its equilibrium position.
At point A $\to$ Slope $\frac{d U}{d x}$ is negative so F is positive
At point C $\to$ Slope is positive so F is negative
At point B $\to$ It is the point of stable equilibrium
At point B $\to$ $U = U_{m i n} \Rightarrow \frac{d U}{d x} = 0$ and $\frac{d ^{2} U}{d x ^{2}} =$ positive

Unstable Equilibrium

After a particle is slightly displaced from its equilibrium position, if If it moves away from the equilibrium position, it is considered to be in unstable equilibrium
At point D $\to$ slope $\frac{d U}{d x}$ is positive so F is negative
At point G $\to$ slope $\frac{d U}{d x}$ is negative so F is positive
At point E $\to$ It is the point of unstable equilibrium
At point E $\to$ $U = U_{m a x} \Rightarrow \frac{d U}{d x} = 0 an d \frac{d ^{2} U}{d x ^{2}} = N e g a t i v e$

Neutral Equilibrium

After a particle is slightly displaced from its equilibrium position. If no forces act on it, the equilibrium is referred to as neutral equilibrium
Point H corresponds to neutral equilibrium $(U = C o n s t an t)$

$\Rightarrow \frac{d U}{d x} = 0, \frac{d ^{2} U}{d x ^{2}} = 0$

6.0Sample Questions On Potential Energy

Q-1. A rectangular rod of mass 'm' and length 'l' is placed in a horizontal position. Find the work done by the external force against gravity to alter its configuration from horizontal position to vertical position.

Solution: Displacement between the positions of center of mass of rod is (in a direction parallel to the force of gravity)

$W = - Change in PE of the rod = m g \frac{l}{2}$

Q-2. The P.E. for a conservative force system is given by U = ax² – bx, where a and b are constants. Find out the (a) expression for force (b) potential energy at equilibrium.

Solution:

(a). For Conservative Force $F = - \frac{d U}{d x} = - (2 a x - b) = - 2 a x + b$

(b). $U = a (\frac{b}{2 a})^{2} - b (\frac{b}{2 a}) = \frac{b ^{2}}{4 a} - \frac{b ^{2}}{2 a} = - \frac{b ^{2}}{4 a}$

Q-3. A body is dropped from height 8 m. After striking the surface it rises to 6 m, what is the fractional loss in kinetic energy during impact? Assuming the frictional resistance to be negligible.

Solution: Fractional loss in kinetic energy =

$\frac{Loss in K E}{Initial K E} = \frac{Loss in PE}{Initial PE} = \frac{m g ( 8 - 6 )}{m g ( 8 )} = \frac{2}{8} = \frac{1}{4}$

Q-4. An object moves in a potential region given by $U = 2 x^{2} - 2 x + 100 J$ . Value of x at which particle will be in equilibrium.

Solution:

$F = - \frac{d U}{d x}$

For Equilibrium,

$\frac{d U}{d x} = 0 \Rightarrow 4 x - 2 = 0$

$x = \frac{2}{4} = 0.5$

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