Potential Energy

 

Table of Content

Illustration Gravitational Potential Energy Relationship between Kinetic Energy and Potential Energy Elastic Potential Energy and Kinetic Energy Conservation of Energy Solved Problems Related Resources

Potential energy is the energy acquired by a body due to change in its position or its configuration. A force acting on a body or a system can also change its potential energy.

Potential energy has its concept associated with conservative forces i.e., the forces for which the work done is independent of the path followed. For example, gravitational force, electrostatic force etc.

Illustration

(a) Let a body raised upwards, against gravitational force, without acceleration. The force applied on the body moves it upwards while gravitational force opposes it. In this case, there is no change in kinetic  energy of the body while it acquires potential energy due to a change in its position.

(b) Let a force be applied to a body which is used to (compress) a spring against a rigid support. Work in this case, is done against elastic forces and is stored in the body in the form of potential energy.

All the forces, explained in this section, are conservative forces. On removal of these forces, the potential energy (stored) can be released into another form of energy. Work done against non-conservative forces (friction, viscous drag etc.) cannot be released back. That work is converted into heat and gets absorbed by surrounding systems.

The conservative force is given by the negative gradient of certain scalar quantity, called potential energy.

So, f = -dU/dx

Thus, 

 

In the absence of an external force, the sum total of kinetic and potential energies always remains constant.

?K + ?U = 0 

Some other examples illustrating the concept of potential energy are given below:

(i) A wire supported by a rigid support and loaded at the other acquires potential energy.

(ii) A beam supported on two knife edges and loaded in the center acquires potential energy.

(iii) Water flowing in a river on a hill has potential energy stored in it. On falling down, potential energy is converted into electric energy in hydraulic dams.

Every substance which is in state of possessing potential energy has a tendency to lose it. That is why a wound up spring tends to open up. A stretched wire, a bent strip or a bent beam tends to acquire original configuration.

Thus, every system in equilibrium tends to exist in the state of minimum potential energy.

This basic fact has found a number of applications to explain the concept of a variety of phenomena in physics. 

Here we shall discuss two cases:

(i) Potential energy of a body due to gravity, called gravitational potential energy.

(ii) Potential energy of a spring when it is elongated or compressed by an external force called elastic potential energy.

Gravitational Potential Energy

It is the energy acquired by a body by virtue of its position in the gravitational field.

A body having mass ‘m’  always acted by a constant force F (= mg).  It is directed towards the earth. Here g is acceleration due to gravity.

Consider a body of mass m moving vertically downward from height y₂ to y₁, as shown in the Figure given below, the work done by this constant force of gravity is given by

 

                W = -mg(y₂-y₁) = - (mgy₂ - mgy₁)

Here W depends on the difference the height or position.

So, we can define gravitational Potential Energy associated with the body as

                U = mgy

Hence W = -(U₂ - U₁) = -ΔU

The negative sign implies that when gravity does positive work, the potential energy decreases. When gravity does negative work (the body moves upward), the potential energy increases.

 We can observe that when a body falls from a height, it accelerates and increases its speed and hence gains K.E. This is at the expense of gravitational P.E. Hence we can relate potential energy and kinetic energy.

Relationship between Kinetic Energy and Potential Energy

We can relate P.E. and K.E. with the help of work energy theorem. Here the external force is gravitational force and the work done, W, by this force is equal to change in K.E. of the body. According Work Energy Theorem this W is equal to change in P.E. i.e.

                W = ΔK = -ΔU    => ΔK = + ΔU = 0

Let the speed of body at height y₁ and y₂ be v₁ and v₂ respectively. (see Figure given below)

                1/3mv₂² - 1/2mv₁² = -mg(y₂ - y₁)

                1/2mv₁² + mgy₁  =  1/2mv₂²+ mgy₁

Therefore the total energy, consists of K.E. and P.E., is conserved.

Now if the force is other than gravity and the work done by these forces is W', then according to Work-Energy theorem,

                 W = ΔK + ΔU

Elastic Potential Energy and Kinetic Energy

When a spring is elongated or compressed, work is done against the elastic restoring force of the spring (which we have discussed in previous illustration, see Figure given below). The work done is stored in the spring as elastic potential energy. Consider a mass 'm' attached to a spring. An applied force P stretches the spring by pulling the mass to the right. As shown  earlier for an applied force P which is proportional to x, the work done is (1/2)kx². This amount of work done is called the elastic potential energy of the body.

The elastic restoring force acts in a direction opposite to the displacement. Hence, the worked one by the restoring force when the body moves x₁ to x₂ is

                W_el = -(1/2)kx₂² -(-(1/2)kx₁²)

                       = -(U₂ - U₁) = -ΔU

Let W' be the work done by the applied force P.

Total work done = change in kinetic energy.

                W' + W_el = ΔK,

                W' = ΔK + ΔU,

                W' = (1/2mv₂² - 1/2mv₁²) + (1/2Kx₂² - 1/2Kx₁²)

                W' = (1/2mv₂² + 1/2mv₁²) - (1/2mv₂²+ 1/2Kx₁²)

As defined earlier, the sum of the kinetic and potential energies of the body is the total mechanical energy. Thus the work of all forces acting on the body with the exception of the elastic force equals the change in total mechanical energy of the body.

If there is no force other than the elastic restoring force acting on the body, then W' = 0, and the mechanical energy of the system is conserved.

               1/2mv₁² + 1/2kx₁² = 1/2mv₂² + 1/2kx₂²

Conservation of Energy

Law of conservation of energy was first stated by Robert Mayor in 1842. It states that,

”Energy can neither be created nor destroyed. It can be converted from one form to another. The sum total of energy, In this universe, is always same.” 

Solved Problems

Problem 1:-

Two snow – covered peaks are at elevations of 862 m and 741 m above the valley between them. A ski run extends from the top of the higher peak to the top of the lower one as shown in below figure. (a) A skier  starts from rest on the higher peak. At what speed would he arrive at the lower peak if he coasts without using the poles? Assume icy conditions, so that there is no friction. (b) After snow fall, a 54.4 kg skier making the same run also without using the poles only just makes it to the lower peak. By how much does the internal energy of her skies and the snow over which she traveled increase?

Concept:-

Kinetic energy K of a body is defined as,

K = ½ mv²

Here m is the mass of the body and v is the velocity of the body.

So, the initial kinetic energy K_i of the body having initial velocity v_i will be,

K_i = ½ mv_i²

And the final kinetic energy K_f of the body having final velocity v_f will be,

K_f = ½ mv_f²

U = mgh

Here m is the mass of the body, g is the free fall acceleration and h is the fall of height.

So, the initial potential energy U_i of the body having initial height y_i will be,

U_i = mg y_i

And the final potential energy U_f of the body having final height y_f will be,

U_f = mg y_f

(a) When the skier starts from rest on the higher peak, the initial velocity is zero results zero initial kinetic energy.

So, K_i = ½ mv_i²

          = ½ m(0)²

          = 0

In accordance to law of conservation of energy, the sum of initial potential and kinetic energy (U_i + K_i) of the skier at the top of the peak is equal to the sum of final potential and kinetic energy (U_f + K_f)  of the skier at the bottom.

So, U_i + K_i = U_f + K_f

U_i + K_i = U_f + K_f

 So, K_f = U_i - U_f

Substitute ½ mv_f² for K_f, mg y_i for U_i and mg y_f for U_f in the equation K_f = U_i - U_f,

K_f = U_i - U_f

 ½ mv_f²   = mg y_i - mg y_f

So, v_f = √2 g (y_i - y_f)

v_f = √2 g (y_i - y_f)

    = √2 (9.81 m/s²) (862 m - 741 m)

   = 48.7 m/s

   = (48.7 m/s) (10^-3 km/1 m)

   = (48.7 m/s) (10^-3 km/1 m) (3600 s/1 h)

   = 175.320 km/h

Rounding off to three significant figures, the speed at which the skier arrives at the lower peak would be 175 km/h.

So, ΔE = U_i - U_f

           = mg y_i - mg y_f

           = mg (y_i – y_f)

To obtain the increase in internal energy ΔE of her skis and the snow over which she travelled, substitute 54.4 kg for mass m, 9.81 m/s² for g, 862 m for y_i and 741 m for y_f in the equation ΔE = mg (y_i – y_f),

ΔE  = mg (y_i – y_f)

      = (54.4 kg) (9.81 m/s²) (862 m - 741 m)

      = - 6.46×10⁴ J

The negative sign signifies that, the internal energy of the snow and skis increased by 6.46×10⁴ J.

From the above observation we conclude that, the increase in internal energy ΔE of her skis and the snow over which she travelled would be 6.46×10⁴ J. 

Problem 2:-

One end of a vertical spring is fastened to the ceiling. A weight is attached to the other end and slowly lowered to its equilibrium position. Show that the loss of gravitational potential energy of the weight equals one-half the gain in spring potential energy.

Concept:-

The potential energy (ΔU_s) of a spring is defined as the,

ΔU_s = ½ ky²      …… (1)

Where k is the spring constant and y is the extension of the spring from its original position.

Gravitational potential energy (ΔU_g) of the weight (mg) would be,

ΔU_g = -mgy        …… (2)

Negative sign indicates that force is acting against the gravity.

Force acting (F) on the spring due to an extension y will be,

F = ky                …… (3)

Where, k is the spring constant.

Weight is equal (mg) to the force due to extension of the spring (F).

So, F = mg     …… (4)

From equation (3) and (4), we get,

F = ky = mg     …… (5)

From equation (1), we get,

ΔU_s = ½ ky²

       = ½ (ky) (y)     …… (6)

ΔU_s= ½ (ky) (y)

      = ½ mgy

So, 2 ΔU_s = mgy

                = - (-mgy)

               = - ΔU_g          (Since, ΔU_g = -mgy)    …… (7)

Negative sign indicates that force is acting against the gravity.

From equation (7) we observed that, the loss of gravitational potential energy of the weight equals one-half the gain in spring potential energy. 

Problem 3:-

A very small ice cube is released  from the edge of a hemispherical frictionless bowl whose radius is 23.6 cm as shown in below figure. How fast is the cube moving at the bottom of the bowl?

Concept:-

In an isolated system in which only conservative forces act, the total mechanical energy remains constant. That is, the initial value of the total mechanical energy (K_i +U_i)  is equal to the final value (K_f +U_f). Here K_i is the initial kinetic energy, U_i is the initial potential energy, K_f is the final kinetic energy and U_f is the final potential energy.

So, K_f +U_f = K_i +U_i

    ½ mv_f² + mgy_f = ½ mv_i² + mgy_i

Since the ice cube is released from the edge of a hemispherical frictionless bowl, so we have to substitute –r for y_f (negative sign due to down ward direction) , 0 for v_i (at the point of release) and 0 for y_i (at the point of release) in the equation ½ mv_f² + mgy_f = ½ mv_i² + mgy_i.

½ mv_f² + mgy_f = ½ mv_i² + mgy_i

½ v_f² + gy_f = ½ v_i² + gy_i

½ v_f² + g(-r) = ½ (0)² + g(0)

½ v_f² = gr

So, the final velocity v_f of the cube at the bottom of the bowl would be,

 v_f = √2gr

v_f = √2gr

    = √2(9.81 m/s²) (23.6 cm)

    = √2(9.81 m/s²) (23.6 cm×10^-2 m/1 cm)

    = 2.15 m/s

From the above observation we conclude that, the speed of the cube at the bottom of the bowl would be 2.15 m/s.

Related Resources

• Click here for the Detailed Syllabus of IIT JEE Physics.

• Look into the Sample Papers of Previous Years to get a hint of the kinds of questions asked in the exam.

• You can get the knowledge of Useful Books of Physics.