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.
\(W = \vec F_C . \Delta \vec r\)
\(W= F_C (- \hat j) . [(x_B - x_A)\hat i+ (y_B- y_A) \hat j]\)
\(W= -[F_C (y_B -y_A)]\)
A 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 = \(\vec F_g . \Delta \vec r\)
\( = 20(-\hat j ) . (2\hat i +\hat j )\)
\(= -20 \,J\)
Option C is Correct
\(U = f(x)\)
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
\(U = f(x)\)
A Stable
B Unstable
C Neutral
D Quasi - stable
Example: A force acting in x-direction with constant magnitude on a body is conservative force.
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 \(\vec F\) 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.
Option B is Correct
\(W_C = - \Delta U\)
i.e., work done by a conservative force is negative of the change in the 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 \(-40\,J \,\)each
B \(-80\,J ,\, -40\,J \)
C \(-60\,J, \,40 \,J\)
D \(-40\,J,\, -50\,J \)
Option A is Correct
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.
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
Kinetic Energy (KE) + Potential Energy (PE) = Constant
\(K.E +P.E = constant\)
Emec = Constant
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 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 \(x = 2\)
\(ME = 40 \,J\)
\(P.E = 0\)
\(KE = ME- PE\)
\(= 40 \,J\)
Option A is Correct
Kinetic Energy (KE) + Potential Energy (PE) = Constant
\(K.E +P.E = constant\)
Emec = Constant
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.
\(W_C = - \Delta U\)...(1)
then \(W_C = F_C . \Delta x\)...(2)
\(F_C. \Delta x = - \Delta U\)
\(F_C = - \dfrac{\Delta U}{\Delta x}\)
\( F_C = \lim\limits_{\Delta x\to 0} \left(-\dfrac{\Delta U}{\Delta x}\right)\)
\( F_C = - \dfrac{dU}{dx}\)
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.
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.
Option C is Correct
Mathematically, \(F_x = - \dfrac{dU}{dx}\)
Also, derivative of \(U\) with respect to \(x\) is the slope of \(U-x\) graph.
then,
\(\dfrac{dU}{dx} >0\,\,\,and\,\,\, F<0\) i.e., force is acting away from B
\(\dfrac{dU}{dx} <0\,\,\,and\,\,\, F>0\) i.e., force is acting away from B
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,
\(\dfrac{dU}{dx} >0\,\,\,and\,\,\, F<0\) i.e., force is acting away from B
\(\Rightarrow\dfrac{dU}{dx} <0\,\,\,and\,\,\, F>0\) i.e., force is acting away from B
Hence, option (A) is correct.
Option A is Correct
