355581
A particle is released from rest at origin. It moves under influence of potential field \(U=x^{2}-3 x\), kinetic energy at \(x=2\) is:
1 2 \(J\)
2 1 \(J\)
3 1.5 \(J\)
4 0 \(J\)
Explanation:
As the total energy is constant \(T.{E_i} = T.{E_f}\) \({K_i} + {U_i} = {K_f} + {U_f}\) \({K_i} = \frac{1}{2}m\left( {{0^2}} \right) = 0\) \({U_i} = 0\) (At origin) \({U_f} = {\left( 2 \right)^2} - 3\left( 2 \right) = - 2J\) \( \Rightarrow {K_f} = 2J\)
PHXI06:WORK ENERGY AND POWER
355582
The following plot shows the variation of potential energy \((U)\) of a system versus postion \((x)\). Predict the wrong option.
1 Point \(D\) is position of neutral equilibrium
2 Point \(B\) is position of unstable equilibrium
3 Point \(\mathrm{C}\) is position of stable equilibrium
4 Point \(A\) is position of neutral equilibrium
Explanation:
At stable equilibrium position \(U\) is minimum, at unstable it is maximum and at neutral position it is constant.
PHXI06:WORK ENERGY AND POWER
355583
An ice cube has been given a push and slides without friction on a level table. Which is correct?
1 It is in neutral equilibrium
2 It is in unstable equilibrium
3 It is in stable equilibrium
4 It is not in equilibrium
Explanation:
The ice cube is in neutral equilibrium. Since it has zero acceleration
PHXI06:WORK ENERGY AND POWER
355584
A particle free to move along \(x\)-axis has potential energy given by \(U(x)=K\left[1-\exp (-x)^{2}\right]\) for \(-\infty \leq x \leq \infty\), where \(\mathrm{K}\) is a positive constant of appropriate dimensions. Then:
1 At point away from the origin, the particle is in unstable equilibrium
2 For any finite non-zero value of \(x\), there is force directed away from the origin
3 If its total mechanical energy is \(K / 2\), it has its minimum \(\mathrm{KE}\) at the origin
4 For small displacement from \(x=0\), the motion is simple harmonic
Explanation:
The graph between \(\mathrm{U} \& x\) is \(\mathrm{U}=\mathrm{K}\left[1-\mathrm{e}^{-x^{2}}\right]=\mathrm{K}\left[1-\dfrac{1}{\mathrm{e}^{x^{2}}}\right]\) A t \(x=0 \quad \mathrm{U}=\mathrm{K}[1-1]=0\) As \(x \rightarrow \infty \quad \mathrm{U}=\mathrm{K}\left[1-\mathrm{e}^{-\infty}\right]=\mathrm{K}\) It is an exponentially increasing graph of potential energy \(U\). From the graph it is clear that at origin. Potential energy \(U\) is minimum (therefore, kinetic energy will be maximum) and force acting on the particle is also zero because \(F=\dfrac{-d U}{d x}=(\) slope of \(U-x\) graph \()=0\) Therefore, origin is the stable equilibrium position. Hence, particle will oscillate simple harmonically about \(x=0\) for small displacements. Therefore, option \(\mathrm{d}\) is correct. (1), (2) and (3) options are wrong due to the following reasons: From the graph we can see that slope is zero at \(x=0\) and \(x= \pm \infty\). Now among these equilibriums stable equilibrium position is that where \(U\) is minimum (Here \(x=0\) ). Unstable equilibrium position is that where \(U\) is maximum (Here none). Neutral equilibrium position is that where \(U\) is constant (here \(x= \pm \infty\) ). Therefore, option (1) is wrong. For any infinite non-zero value of \(x\), force is directed towards the origin because origin is in stable equilibrium position. Therefore, option (2) is incorrect. At the origin, potential energy is minimum, hence kinetic energy will be maximum. Therefore, option (3) is also wrong.
355581
A particle is released from rest at origin. It moves under influence of potential field \(U=x^{2}-3 x\), kinetic energy at \(x=2\) is:
1 2 \(J\)
2 1 \(J\)
3 1.5 \(J\)
4 0 \(J\)
Explanation:
As the total energy is constant \(T.{E_i} = T.{E_f}\) \({K_i} + {U_i} = {K_f} + {U_f}\) \({K_i} = \frac{1}{2}m\left( {{0^2}} \right) = 0\) \({U_i} = 0\) (At origin) \({U_f} = {\left( 2 \right)^2} - 3\left( 2 \right) = - 2J\) \( \Rightarrow {K_f} = 2J\)
PHXI06:WORK ENERGY AND POWER
355582
The following plot shows the variation of potential energy \((U)\) of a system versus postion \((x)\). Predict the wrong option.
1 Point \(D\) is position of neutral equilibrium
2 Point \(B\) is position of unstable equilibrium
3 Point \(\mathrm{C}\) is position of stable equilibrium
4 Point \(A\) is position of neutral equilibrium
Explanation:
At stable equilibrium position \(U\) is minimum, at unstable it is maximum and at neutral position it is constant.
PHXI06:WORK ENERGY AND POWER
355583
An ice cube has been given a push and slides without friction on a level table. Which is correct?
1 It is in neutral equilibrium
2 It is in unstable equilibrium
3 It is in stable equilibrium
4 It is not in equilibrium
Explanation:
The ice cube is in neutral equilibrium. Since it has zero acceleration
PHXI06:WORK ENERGY AND POWER
355584
A particle free to move along \(x\)-axis has potential energy given by \(U(x)=K\left[1-\exp (-x)^{2}\right]\) for \(-\infty \leq x \leq \infty\), where \(\mathrm{K}\) is a positive constant of appropriate dimensions. Then:
1 At point away from the origin, the particle is in unstable equilibrium
2 For any finite non-zero value of \(x\), there is force directed away from the origin
3 If its total mechanical energy is \(K / 2\), it has its minimum \(\mathrm{KE}\) at the origin
4 For small displacement from \(x=0\), the motion is simple harmonic
Explanation:
The graph between \(\mathrm{U} \& x\) is \(\mathrm{U}=\mathrm{K}\left[1-\mathrm{e}^{-x^{2}}\right]=\mathrm{K}\left[1-\dfrac{1}{\mathrm{e}^{x^{2}}}\right]\) A t \(x=0 \quad \mathrm{U}=\mathrm{K}[1-1]=0\) As \(x \rightarrow \infty \quad \mathrm{U}=\mathrm{K}\left[1-\mathrm{e}^{-\infty}\right]=\mathrm{K}\) It is an exponentially increasing graph of potential energy \(U\). From the graph it is clear that at origin. Potential energy \(U\) is minimum (therefore, kinetic energy will be maximum) and force acting on the particle is also zero because \(F=\dfrac{-d U}{d x}=(\) slope of \(U-x\) graph \()=0\) Therefore, origin is the stable equilibrium position. Hence, particle will oscillate simple harmonically about \(x=0\) for small displacements. Therefore, option \(\mathrm{d}\) is correct. (1), (2) and (3) options are wrong due to the following reasons: From the graph we can see that slope is zero at \(x=0\) and \(x= \pm \infty\). Now among these equilibriums stable equilibrium position is that where \(U\) is minimum (Here \(x=0\) ). Unstable equilibrium position is that where \(U\) is maximum (Here none). Neutral equilibrium position is that where \(U\) is constant (here \(x= \pm \infty\) ). Therefore, option (1) is wrong. For any infinite non-zero value of \(x\), force is directed towards the origin because origin is in stable equilibrium position. Therefore, option (2) is incorrect. At the origin, potential energy is minimum, hence kinetic energy will be maximum. Therefore, option (3) is also wrong.
355581
A particle is released from rest at origin. It moves under influence of potential field \(U=x^{2}-3 x\), kinetic energy at \(x=2\) is:
1 2 \(J\)
2 1 \(J\)
3 1.5 \(J\)
4 0 \(J\)
Explanation:
As the total energy is constant \(T.{E_i} = T.{E_f}\) \({K_i} + {U_i} = {K_f} + {U_f}\) \({K_i} = \frac{1}{2}m\left( {{0^2}} \right) = 0\) \({U_i} = 0\) (At origin) \({U_f} = {\left( 2 \right)^2} - 3\left( 2 \right) = - 2J\) \( \Rightarrow {K_f} = 2J\)
PHXI06:WORK ENERGY AND POWER
355582
The following plot shows the variation of potential energy \((U)\) of a system versus postion \((x)\). Predict the wrong option.
1 Point \(D\) is position of neutral equilibrium
2 Point \(B\) is position of unstable equilibrium
3 Point \(\mathrm{C}\) is position of stable equilibrium
4 Point \(A\) is position of neutral equilibrium
Explanation:
At stable equilibrium position \(U\) is minimum, at unstable it is maximum and at neutral position it is constant.
PHXI06:WORK ENERGY AND POWER
355583
An ice cube has been given a push and slides without friction on a level table. Which is correct?
1 It is in neutral equilibrium
2 It is in unstable equilibrium
3 It is in stable equilibrium
4 It is not in equilibrium
Explanation:
The ice cube is in neutral equilibrium. Since it has zero acceleration
PHXI06:WORK ENERGY AND POWER
355584
A particle free to move along \(x\)-axis has potential energy given by \(U(x)=K\left[1-\exp (-x)^{2}\right]\) for \(-\infty \leq x \leq \infty\), where \(\mathrm{K}\) is a positive constant of appropriate dimensions. Then:
1 At point away from the origin, the particle is in unstable equilibrium
2 For any finite non-zero value of \(x\), there is force directed away from the origin
3 If its total mechanical energy is \(K / 2\), it has its minimum \(\mathrm{KE}\) at the origin
4 For small displacement from \(x=0\), the motion is simple harmonic
Explanation:
The graph between \(\mathrm{U} \& x\) is \(\mathrm{U}=\mathrm{K}\left[1-\mathrm{e}^{-x^{2}}\right]=\mathrm{K}\left[1-\dfrac{1}{\mathrm{e}^{x^{2}}}\right]\) A t \(x=0 \quad \mathrm{U}=\mathrm{K}[1-1]=0\) As \(x \rightarrow \infty \quad \mathrm{U}=\mathrm{K}\left[1-\mathrm{e}^{-\infty}\right]=\mathrm{K}\) It is an exponentially increasing graph of potential energy \(U\). From the graph it is clear that at origin. Potential energy \(U\) is minimum (therefore, kinetic energy will be maximum) and force acting on the particle is also zero because \(F=\dfrac{-d U}{d x}=(\) slope of \(U-x\) graph \()=0\) Therefore, origin is the stable equilibrium position. Hence, particle will oscillate simple harmonically about \(x=0\) for small displacements. Therefore, option \(\mathrm{d}\) is correct. (1), (2) and (3) options are wrong due to the following reasons: From the graph we can see that slope is zero at \(x=0\) and \(x= \pm \infty\). Now among these equilibriums stable equilibrium position is that where \(U\) is minimum (Here \(x=0\) ). Unstable equilibrium position is that where \(U\) is maximum (Here none). Neutral equilibrium position is that where \(U\) is constant (here \(x= \pm \infty\) ). Therefore, option (1) is wrong. For any infinite non-zero value of \(x\), force is directed towards the origin because origin is in stable equilibrium position. Therefore, option (2) is incorrect. At the origin, potential energy is minimum, hence kinetic energy will be maximum. Therefore, option (3) is also wrong.
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI06:WORK ENERGY AND POWER
355581
A particle is released from rest at origin. It moves under influence of potential field \(U=x^{2}-3 x\), kinetic energy at \(x=2\) is:
1 2 \(J\)
2 1 \(J\)
3 1.5 \(J\)
4 0 \(J\)
Explanation:
As the total energy is constant \(T.{E_i} = T.{E_f}\) \({K_i} + {U_i} = {K_f} + {U_f}\) \({K_i} = \frac{1}{2}m\left( {{0^2}} \right) = 0\) \({U_i} = 0\) (At origin) \({U_f} = {\left( 2 \right)^2} - 3\left( 2 \right) = - 2J\) \( \Rightarrow {K_f} = 2J\)
PHXI06:WORK ENERGY AND POWER
355582
The following plot shows the variation of potential energy \((U)\) of a system versus postion \((x)\). Predict the wrong option.
1 Point \(D\) is position of neutral equilibrium
2 Point \(B\) is position of unstable equilibrium
3 Point \(\mathrm{C}\) is position of stable equilibrium
4 Point \(A\) is position of neutral equilibrium
Explanation:
At stable equilibrium position \(U\) is minimum, at unstable it is maximum and at neutral position it is constant.
PHXI06:WORK ENERGY AND POWER
355583
An ice cube has been given a push and slides without friction on a level table. Which is correct?
1 It is in neutral equilibrium
2 It is in unstable equilibrium
3 It is in stable equilibrium
4 It is not in equilibrium
Explanation:
The ice cube is in neutral equilibrium. Since it has zero acceleration
PHXI06:WORK ENERGY AND POWER
355584
A particle free to move along \(x\)-axis has potential energy given by \(U(x)=K\left[1-\exp (-x)^{2}\right]\) for \(-\infty \leq x \leq \infty\), where \(\mathrm{K}\) is a positive constant of appropriate dimensions. Then:
1 At point away from the origin, the particle is in unstable equilibrium
2 For any finite non-zero value of \(x\), there is force directed away from the origin
3 If its total mechanical energy is \(K / 2\), it has its minimum \(\mathrm{KE}\) at the origin
4 For small displacement from \(x=0\), the motion is simple harmonic
Explanation:
The graph between \(\mathrm{U} \& x\) is \(\mathrm{U}=\mathrm{K}\left[1-\mathrm{e}^{-x^{2}}\right]=\mathrm{K}\left[1-\dfrac{1}{\mathrm{e}^{x^{2}}}\right]\) A t \(x=0 \quad \mathrm{U}=\mathrm{K}[1-1]=0\) As \(x \rightarrow \infty \quad \mathrm{U}=\mathrm{K}\left[1-\mathrm{e}^{-\infty}\right]=\mathrm{K}\) It is an exponentially increasing graph of potential energy \(U\). From the graph it is clear that at origin. Potential energy \(U\) is minimum (therefore, kinetic energy will be maximum) and force acting on the particle is also zero because \(F=\dfrac{-d U}{d x}=(\) slope of \(U-x\) graph \()=0\) Therefore, origin is the stable equilibrium position. Hence, particle will oscillate simple harmonically about \(x=0\) for small displacements. Therefore, option \(\mathrm{d}\) is correct. (1), (2) and (3) options are wrong due to the following reasons: From the graph we can see that slope is zero at \(x=0\) and \(x= \pm \infty\). Now among these equilibriums stable equilibrium position is that where \(U\) is minimum (Here \(x=0\) ). Unstable equilibrium position is that where \(U\) is maximum (Here none). Neutral equilibrium position is that where \(U\) is constant (here \(x= \pm \infty\) ). Therefore, option (1) is wrong. For any infinite non-zero value of \(x\), force is directed towards the origin because origin is in stable equilibrium position. Therefore, option (2) is incorrect. At the origin, potential energy is minimum, hence kinetic energy will be maximum. Therefore, option (3) is also wrong.
