Informative line

Conservative Force And Conservation Of Mechanical Energy

Learn principle of conservation of mechanical energy and conservative forces with examples, practice to find relation between work done by a conservative force (wc) and potential energy (u) and calculate conservation of mechanical energy from potential energy (u) curve.

Calculation of Work Done by Conservative Forces

• The work done by a  conservative force on a particle, moving between any two points, is independent of the  path of the particle.
• The work done depends only on the initial and final position of the particle, as shown in figure.
• Consider a particle moves from point A to point B along an arbitrary path under the action of a conservative force, FC?.
• Thus, work done by a conservative force is

$$W = \vec F_C . \Delta \vec r$$

$$W= F_C (- \hat j) . [(x_B - x_A)\hat i+ (y_B- y_A) \hat j]$$

$$W= -[F_C (y_B -y_A)]$$

A particle of mass m= 2 kg is moving from point A to B under the action of gravity, as shown in figure. Find the work done by the force of gravity.

A 40 J

B 10 J

C – 20 J

D 30 J

×

Force of gravity is a conservative force.

Thus, work done by the force of gravity is independent of the path of the particle.

$$\Delta \vec r = (4-2)\hat i + (3-2)\hat j$$

$$= 2 \hat i + \hat j$$

$$\vec F_g = mg(- \hat j)$$

$$= 2 × 10 (-\hat j)$$

$$= 20 (-\hat j)$$

Work done = $$\vec F_g . \Delta \vec r$$

$$= 20(-\hat j ) . (2\hat i +\hat j )$$

$$= -20 \,J$$

A particle of mass m= 2 kg is moving from point A to B under the action of gravity, as shown in figure. Find the work done by the force of gravity.

A

40 J

.

B

10 J

C

– 20 J

D

30 J

Option C is Correct

Mathematical Method to Check Nature of Equilibrium

• Let potential energy U is a function of position $$x$$

$$U = f(x)$$

• Calculate $$\dfrac{dU}{dx}$$ and put  $$\dfrac{dU}{dx} =0$$ to find  values of $$x$$
• Calculate $$\dfrac{d^2U}{dx^2}$$ and put values of $$x$$ in $$\dfrac{d^2U}{dx^2}$$.
• If $$\dfrac{d^2U}{dx^2} < 0,$$ then it gives maxima of the curve, i.e., point of unstable equilibrium.
• If $$\dfrac{d^2U}{dx^2} > 0,$$ then it gives minima of the curve, i.e., point of stable equilibrium.
• If $$\dfrac{d^2U}{dx^2} = 0,$$ then it gives point of neutral equilibrium.
• Here, points A,C are points of unstable equilibrium.
• Points B,D are points of stable equilibrium.
• Point E is point of neutral equilibrium.

If potential energy U is a function of position $$x$$, then choose the correct option for position $$x =a$$.

A If $$\dfrac{dU}{dx} =0$$ ; $$x =a$$ is a point of equilibrium

B If $$\dfrac{d^2U}{dx^2} >0$$ ; $$x =a$$ is a point of unstable equilibrium

C If $$\dfrac{d^2U}{dx^2} <0$$ ; $$x =a$$ is a point of stable equilibrium

D If $$\dfrac{d^2U}{dx^2} =0$$ ; $$x =a$$ is a point of stable equilibrium

×

At any value of $$x ,$$ if  $$\dfrac{dU}{dx} = 0$$

that value of $$x$$ is a point of equilibrium.

At that value, if $$\dfrac{d^2U}{dx^2} < 0,$$ then it gives maxima of the curve, i.e., point of unstable equilibrium.

If $$\dfrac{d^2U}{dx^2} > 0,$$ then it gives minima of the curve, i.e., point of stable equilibrium.

If $$\dfrac{d^2U}{dx^2} = 0,$$ then it gives point of neutral equilibrium.

Hence, option (A) is correct.

If potential energy U is a function of position $$x$$, then choose the correct option for position $$x =a$$.

A

If $$\dfrac{dU}{dx} =0$$ ; $$x =a$$ is a point of equilibrium

.

B

If $$\dfrac{d^2U}{dx^2} >0$$ ; $$x =a$$ is a point of unstable equilibrium

C

If $$\dfrac{d^2U}{dx^2} <0$$ ; $$x =a$$ is a point of stable equilibrium

D

If $$\dfrac{d^2U}{dx^2} =0$$ ; $$x =a$$ is a point of stable equilibrium

Option A is Correct

Determining the Nature of Equilibrium at Given Point

• Let potential energy U is a function of position $$x$$

$$U = f(x)$$

• Calculate $$\dfrac{dU}{dx}$$ and put  $$\dfrac{dU}{dx} =0$$ to find values of $$x$$
• Calculate $$\dfrac{d^2U}{dx^2}<0$$ and put values of $$x$$ in $$\dfrac{d^2U}{dx^2}$$.
• If $$\dfrac{d^2U}{dx^2} < 0,$$ then it gives maxima of the curve, i.e., point of unstable equilibrium.
• If $$\dfrac{d^2U}{dx^2} > 0,$$ then it gives minima of the curve, i.e., point of stable equilibrium.
• If $$\dfrac{d^2U}{dx^2} = 0,$$ then it gives point of neutral equilibrium.
• Here, points A, C are points of unstable equilibrium.
• Points B, D are points of stable equilibrium.
• Point E is point of neutral equilibrium.

If potential energy is given as a function of $$x$$, $$U = x^3 -3x^2$$. Find the nature of equilibrium at $$x = 2$$.

A Stable

B Unstable

C Neutral

D Quasi - stable

×

Potential energy, $$U = x^3 -3x^2$$

$$\Rightarrow\dfrac{dU}{dx} = 3x^2 - 6x$$

Put  $$\dfrac{dU}{dx} =0$$

$$3x^2 -6x =0$$

$$3x(x-2) =0$$

$$x = 0,\,2$$

So, $$x = 0,\,2$$ are the points of equilibrium.

$$\Rightarrow \dfrac{d^2U}{dx^2} = 6x-6$$

$$\Rightarrow \left(\dfrac{d^2U}{dx^2}\right)_{x=2} = +6$$

Since, $$\dfrac{d^2U}{dx^2 }>0$$

Therefore, $$x = 2$$ is the point of stable equilibrium.

If potential energy is given as a function of $$x$$, $$U = x^3 -3x^2$$. Find the nature of equilibrium at $$x = 2$$.

A

Stable

.

B

Unstable

C

Neutral

D

Quasi - stable

Option A is Correct

Conservative Forces

• There are many kinds of conservative forces among which two are given here.
• The forces which act in a constant direction with constant magnitude are conservative forces.

Example: A force acting in x-direction with constant magnitude on a body is conservative force.

• The forces which always act towards a point with constant magnitude or it depends on distance from the point are also conservative forces.

Example: Gravitational force is a conservative force as it always act towards center of earth and it depends upon the distance from the center of earth.

A particle is moved from $$Q$$ to $$P$$ along the path shown. A force $$\vec F$$ is applied on it in such a way that $$\vec F$$ is constant in magnitude and always along $$-x-$$ axis. Which one of the following options is correct?

A $$\vec F$$ is non-conservative

B $$\vec F$$ is conservative

C Cannot be determined

D

×

At point $$1,\;\vec F$$ is in $$(-x)-$$ axis.

At point $$2,\;\vec F$$ is in $$(-x)-$$ axis.

$$\vec F$$ is constant both in magnitude as well as in direction.

Thus, $$\vec F$$ is conservative.

A particle is moved from $$Q$$ to $$P$$ along the path shown. A force $$\vec F$$ is applied on it in such a way that $$\vec F$$ is constant in magnitude and always along $$-x-$$ axis. Which one of the following options is correct?

A

$$\vec F$$ is non-conservative

.

B

$$\vec F$$ is conservative

C

Cannot be determined

D

Option B is Correct

Relation between Work Done by a Conservative  Force (WC) and Potential Energy (U)

$$W_C = - \Delta U$$

i.e., work done by a conservative force is negative of the change in the potential energy.

• Work done by a conservative force is stored in the system in the form of potential energy.

$$W_C = -\Delta U$$

$$W_C = - (U_{final} - U_{initial})$$

$$W_C = ( U_{initial}-U_{final})$$

$$W_C = U_i - U_f$$

Note :

(1) Since, work done by a conservative force doesn't depend upon the path, therefore change in potential energy will also not depend upon the path.

(2) Negative of the change of potential energy is the work done by a conservative force.

A particle is moving from A to B under the action of a conservative force along two paths (1) and (2). Calculate the work done along both the paths.

A $$-40\,J \,$$each

B $$-80\,J ,\, -40\,J$$

C $$-60\,J, \,40 \,J$$

D $$-40\,J,\, -50\,J$$

×

Since, the force is conservative, thus work done along both the paths will be same.

Hence,

$$W_{C(1)} = W_{C(2)} = W_C$$

$$U_i = U_A = 40 \,J$$

$$U_f = U_B = 80 \,J$$

$$W_C = U_i - U_f$$

$$= (40 -80 )\,J$$

$$= -40 \,J$$

A particle is moving from A to B under the action of a conservative force along two paths (1) and (2). Calculate the work done along both the paths.

A

$$-40\,J \,$$each

.

B

$$-80\,J ,\, -40\,J$$

C

$$-60\,J, \,40 \,J$$

D

$$-40\,J,\, -50\,J$$

Option A is Correct

Visual Idea of Stable, Unstable and Neutral Equilibrium

• Consider a smooth track in a vertical plane, as shown in figure.
• A ball on this track can be balanced at point A, point B and point C (which is any point in region R).
• When the ball is displaced to a point near the equilibrium points, there can be three possible cases.
• Case (I)

The ball does not return to the equilibrium point, thus, it is known as point of unstable equilibrium.

e.g. Point A in figure.

Case (II)

The ball returns back to the equilibrium point, thus, it is known as point of stable equilibrium.

e.g. Point B in figure.

Case (III)

The ball remains at the displaced position. This is also an equilibrium point. Thus, it is known as point of neutral equilibrium.

e.g. Point C in figure.

Choose the incorrect statement.

A A is a point of unstable equilibrium

B B is a point of unstable equilibrium

C C is a point of neutral equilibrium

D P is a point of neutral equilibrium

×

A ball on this track can be balanced at point A, point B, point C and point P.

For point A, the ball returns back to the equilibrium point, thus, it is known as point of stable equilibrium.

For point B, the ball does not return to the equilibrium point, thus, it is known as point of unstable equilibrium.

For point C and P, the ball remains at the displaced position. This is also an equilibrium point. Thus, it is known as point of neutral equilibrium.

Hence, option (A) is incorrect.

Choose the incorrect statement.

A

A is a point of unstable equilibrium

.

B

B is a point of unstable equilibrium

C

C is a point of neutral equilibrium

D

P is a point of neutral equilibrium

Option A is Correct

Conservation of Mechanical Energy from Potential Energy(U) Curve

• When work is done only by conservative internal forces, then mechanical energy of the system remains constant.

Kinetic Energy (KE) + Potential Energy (PE) = Constant

$$K.E +P.E = constant$$

• As mechanical energy, $$E_{mec} = K.E +P.E$$

Emec = Constant

• This result is known as "Principle of Conservation of Mechanical Energy".

Note:-

(1)  If a particle is released from rest or starts motion from a point then, kinetic energy at that point is zero.

(2) Potential energy may be negative but kinetic energy is always non - negative.

A particle is acted upon by a conservative force only. Particle is released from rest at $$x = 1$$. Find the kinetic energy at $$x = 2$$.

A 40 J

B –80 J

C 50 J

D 60 J

×

Since, force is conservative so, mechanical energy is conserved.

Total $$M.E = KE+PE$$

= Constant

At $$x = 1, \,KE =0,\,PE = 40 \,J$$

$$\Rightarrow ME = KE+PE$$

$$= 40 \,J$$

At $$x = 2$$

$$ME = 40 \,J$$

$$P.E = 0$$

$$KE = ME- PE$$

$$= 40 \,J$$

A particle is acted upon by a conservative force only. Particle is released from rest at $$x = 1$$. Find the kinetic energy at $$x = 2$$.

A

40 J

.

B

–80 J

C

50 J

D

60 J

Option A is Correct

Conservation of Mechanical Energy

• When work is done only by conservative internal forces, then mechanical energy of the system remains constant.

Kinetic Energy (KE) + Potential Energy (PE) = Constant

$$K.E +P.E = constant$$

• As mechanical energy, $$E_{mec} = K.E +P.E$$

Emec = Constant

• This result is known as "Principle of Conservation of Mechanical Energy".

Note:-

(1)  If a particle is released from rest or starts motion from a point then, kinetic energy at that point is zero.

(2)  Potential energy may be negative but kinetic energy is always non - negative.

A particle is moving under the action of a conservative force only. Its potential energy is given as $$U = x^2 -4x$$. If it is released from origin with kinetic energy = $$40 \,J$$. Find the kinetic energy at equilibrium position.

A 40 J

B 44 J

C 50 J

D 80 J

×

Calculation of equilibrium position

$$U = x^2 - 4\,x$$

$$\Rightarrow \dfrac{dU}{dx} = 2x-4$$

$$\Rightarrow \dfrac{dU}{dx} =0$$

$$\Rightarrow 2x-4 =0$$

$$x = 2$$

$$x =2$$ is equilibrium position

Calculation of K.E

Total Mechanical Energy (ME) = Kinetic Energy (KE) + Potential Energy (PE)

= Constant

At $$x = 0,\,KE = 40\,J, \,PE =0$$

$$ME =PE +KE$$

$$ME = 40 \,J$$

At  $$x = 2$$

$$PE = (2)^2 - 4(2)$$

$$= -4 \,J$$

$$ME = 40 \,J$$

$$KE = ME - PE$$

$$= 44\,J$$

A particle is moving under the action of a conservative force only. Its potential energy is given as $$U = x^2 -4x$$. If it is released from origin with kinetic energy = $$40 \,J$$. Find the kinetic energy at equilibrium position.

A

40 J

.

B

44 J

C

50 J

D

80 J

Option B is Correct

Point of Stable, Unstable and Neutral Equilibrium in U-x Curve

• Relation between the work done by conservative force $$(W_C)$$ and potential energy (U) is given by

$$W_C = - \Delta U$$...(1)

• If $$F_C$$ is a conservative force and  $$\Delta x$$ is the small displacement of an object, while being acted on by $$F_C$$

then  $$W_C = F_C . \Delta x$$...(2)

•    From equation (1) and (2),

$$F_C. \Delta x = - \Delta U$$

$$F_C = - \dfrac{\Delta U}{\Delta x}$$

• In the limit, $$\Delta x \to 0$$

$$F_C = \lim\limits_{\Delta x\to 0} \left(-\dfrac{\Delta U}{\Delta x}\right)$$

$$F_C = - \dfrac{dU}{dx}$$

• Consider $$U - x$$  curve, as shown in figure.
• If B is a point of stable equilibrium then,

At A, (a point left to B)

$$\dfrac{dU}{dx} <0$$

$$\Rightarrow F>0 \,\,\,\text {as} \,\,\, F= -\dfrac{dU}{dx}$$

i.e., force will act towards B

At C, (a point right to B)

$$\dfrac{dU}{dx} >0$$  $$\Rightarrow F<0$$

i.e., force will act towards B

It can be concluded that sign of  $$\dfrac{dU}{dx}$$ changes from negative to positive.

If B is the point of stable equilibrium in the $$U-x$$ curve, as shown in figure, then choose the correct option.

A At point A, $$\dfrac{dU}{dx} <0$$ and force $$F<0$$

B At point A, $$\dfrac{dU}{dx} >0$$ and force $$F>0$$

C At point C, $$\dfrac{dU}{dx} >0$$ and force $$F<0$$

D At point C, $$\dfrac{dU}{dx} <0$$ and force $$F<0$$

×

If B is a point of equilibrium then,

At A, (a point left to B)

$$\dfrac{dU}{dx} <0$$

$$\Rightarrow F>0 \,\,\,\text {as} \,\,\, F= -\dfrac{dU}{dx}$$

i.e., force will act towards B

At C, (a point right to B)

$$\dfrac{dU}{dx} >0$$  $$\Rightarrow F<0$$

i.e., force will act towards B

Hence, option (C) is correct.

If B is the point of stable equilibrium in the $$U-x$$ curve, as shown in figure, then choose the correct option.

A

At point A, $$\dfrac{dU}{dx} <0$$ and force $$F<0$$

.

B

At point A, $$\dfrac{dU}{dx} >0$$ and force $$F>0$$

C

At point C, $$\dfrac{dU}{dx} >0$$ and force $$F<0$$

D

At point C, $$\dfrac{dU}{dx} <0$$ and force $$F<0$$

Option C is Correct

Point of Stable and Unstable Equilibrium in U-x Curve

• Consider $$U-x$$ curve, as shown in figure.

• For the $$U-x$$  curve, B is the point of unstable equilibrium. We know that,

Mathematically, $$F_x = - \dfrac{dU}{dx}$$

Also, derivative of $$U$$ with respect to $$x$$ is the slope of $$U-x$$ graph.

• If B is the point of unstable equilibrium,

then,

• At A (a point left to B)

$$\dfrac{dU}{dx} >0\,\,\,and\,\,\, F<0$$  i.e., force is acting away from B

• At C (a point right to B )

$$\dfrac{dU}{dx} <0\,\,\,and\,\,\, F>0$$  i.e., force is acting away from B

• Sign of $$\dfrac{dU}{dx}$$ changes from positive to negative.

In the $$U-x$$ curve, as shown in figure, if B is a point of unstable equilibrium, then which one of the following options is correct?

A Force at point C is positive ; $$\dfrac{dU}{dx} <0$$

B Force at point A is negative ; $$\dfrac{dU}{dx} <0$$

C  $$\dfrac{dU}{dx}$$ at C is negative ; $$Force >0$$

D  $$\dfrac{dU}{dx}$$ at A is positive ; $$Force = constant$$

×

If B is the point of unstable equilibrium,

Then,

• At A (a point left to B)

$$\dfrac{dU}{dx} >0\,\,\,and\,\,\, F<0$$  i.e., force is acting away from B

• At C (a point right to B)

$$\Rightarrow\dfrac{dU}{dx} <0\,\,\,and\,\,\, F>0$$  i.e., force is acting away from B

Hence, option (A) is correct.

In the $$U-x$$ curve, as shown in figure, if B is a point of unstable equilibrium, then which one of the following options is correct?

A

Force at point C is positive ; $$\dfrac{dU}{dx} <0$$

.

B

Force at point A is negative ; $$\dfrac{dU}{dx} <0$$

C

$$\dfrac{dU}{dx}$$ at C is negative ; $$Force >0$$

D

$$\dfrac{dU}{dx}$$ at A is positive ; $$Force = constant$$

Option A is Correct