Conservative Force as Gradient of Potential
Trending Questions
Q. A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to (−kr2) where k is constant. What is the total energy of the particle?
- -K/2r2.
- –K2/2r.
- -K/r.
- -K/2r.
Q. The potential energy of 1 kg particle free to move along the X− axis is given by U=(x44−x22) J. The total mechanical energy of the particle is 2 J. Then maximum speed of the particle is (in m/s)
- 3√2
- 1√2
- √2
- 2
Q. The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure.
The potential energy \(U(x)\) versus time \((t)\) plot of the particle is correctly shown in figure :
O &
The potential energy \(U(x)\) versus time \((t)\) plot of the particle is correctly shown in figure :
O &
Q. The potential energy of a particle executing SHM changes from maximum to minimum is 5 sec. Then the time period of SHM is
- 5 s
- 10 s
- 15 s
- 20 s
Q. A particle, which is constrained to move along the x - axis the x - axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x)=−kx+ax3. Here k and a are positive energy U(x) of the particle is
Q.
The potential energy of a particle of mass m at a distance r from a fixed point O is given by V(r)=kr2/2 , where k is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius R about the point O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is (are) true?
- v=√k2m R
- v=√km R
- L=√mk R2
- L=√mk2 R2
Q. The potential energy of a body is given by, U = A − Bx2(Where x is the displacement). The magnitude of force acting on the particle is
- Constant
- Proportional to x
- Inversely proportional to x
- Proportional to x2
Q. The potential energy function of a particle is given by U=−(x2+y2+z2) J, where x, y and z are in meters. Find the force acting on the particle at point A(1 m, 3 m, 5 m).
- √136 N
- √8 N
- √140 N
- 10√14 N
Q. The potential energy of a 4 kg particle free to move along the x− axis is given by U(x)=x33−5x22+6x+3. Total mechanical energy of the particle is 17 J. Then the maximum kinetic energy is
- 10 J
- 2 J
- 0.5 J
- 9.5 J
Q. The figure below shows a graph of potential energy U(x) verses position x for a particle executing one dimentional motion along the x-axis. The total mechanical energy of the system is indicated by the dashed line. Choose the correct statement for the position of particle varrying between points A and G.
- The magnitude of force is maximum at D
- The kinetic energy is maximum at B
- The velocity is zero at A and G
- None of the above
Q. Potential energy U(x) and associated force F(x) bear the relation F(x)=−dU(x)dx. Dependence of potential energy of a two-particle system on the separation x between them is shown in the following figure.
Which of the following graphs shows the correct variation of force F(x) with x?
Which of the following graphs shows the correct variation of force F(x) with x?
Q. The potential energy for a force field →F is given by U(x, y)=cos(x+y). The force acting on a particle at the position given by coordinates (0, π4) is
- 1√2(^i+^j)
- (12^i+√32^j)
- (12^i−√32^j)
- −1√2(^i+^j)
Q. A particle moves in one dimension in a conservative force field. The variation of potential energy with distance is as shown in the graph below. If the particle starts to move from rest from the point A. Then
- The speed is zero at the points A and E
- The acceleration vanishes at the points A, B, C, D, E
- The acceleration vanishes at the points B, C, D
- The speed is maximum at the point D
Q. The potential energy of a particle that is free to move along (x−axis) is given by U=(x3−x22)J, where x is in 'm'. The particle is initially at x=0. Find the magnitude of force at x=2 m.
- 6 N
- 2 N
- 8 N
- 10 N
Q. A fly wheel is in the form of a uniform circular disc of radius 1 m and mass 2 kg. The work which must be done on it to increase its frequency of rotation from 5 rev/s to 10 rev/s is approximately
- 1.5×102 J
- 3×102 J
- 1.5×103 J
- 3×103 J
Q. A satellite is in a circular equatorial orbit of radius 7000 km around the earth. If it is transferred to a circular orbit of double the radius.
Which of the following option has the correct combination considering column-I and column-II.
Column I | Column -II |
(A) angular momentum | (P) increases |
(B) area of earth covered by satellite signal | (Q) decreases |
(C) potential energy | (R) becomes double |
(D) kinetic energy | (S) becomes half |
Which of the following option has the correct combination considering column-I and column-II.
- D→R, S
- A→P
- B→Q, S
- C→S
Q. A particle is moving in a circular path of radius a with constant speed under the action of an attractive conservative force. Potential energy of the particle is given by the relation U=−K2r2, where r is the radial distance of the particle from the centre of the circular path. Its total energy will be:
- K2a2
- Zero
- −32Ka2
- −K4a2
Q. A particle of mass 5 kg moving in the X−Y plane has its potential energy given by U=(−7x+24y) J. The particle is initially at the origin and has a velocity, u=(14.4^i+4.2^j) m/s. Then,
- The particle has speed 20 m/s at t=4 sec
- The particle has an acceleration 25 m/s2
- The acceleration of the particle is normal to the initial velocity
- None of the above are correct
Q. A particle has potential energy dependent on its position on the x axis, represented by the function U(x)=e2x+1 for all real values of x, where U(x) and x are given in standard units. The force it feels at position x=1 is closest to
[Take e=2.72]
[Take e=2.72]
- −8.39 N
- 14.8 N
- −14.8 N
- 8.39 N
Q. The potential energy for a conservative force system is given by U=ax3−bx, where a and b are constants. Choose from the options, the correct x coordinate(s) of the equilibrium position(s).
- +√b2a
- −√b2a
- +√b3a
- −√b3a
Q. A small sphere B of mass 1 kg is released from rest in the position shown and swings freely in a vertical plane, first about O and then about the peg A after the cord comes in contact with the peg. The tension in the cord just after it comes in contact with the peg is
[Take g=10 m/s2]
[Take g=10 m/s2]
- 30 N
- 20 N
- 15 N
- 25 N
Q. A particle is released from rest at origin. It moves under the influence of a potential field, U=x2−3x. Find the Kinetic energy of the particle at x=2.
- 1 J
- 1.5 J
- 0 J
- 2 J
Q. A plot of potential energy function U(x)=kx2, where x is the displacement and k= constant is shown below. The correct conservative force function F(x) is
Q. The potential energy of a certain particle is given by U=12(x2−y2). The force on it is:
- −x^i−y^j
- x^i−y^j
- −x^i+y^j
- −x^i+z^k
Q. The force exerted by a stretched chord at given displacements is shown in the table given below. Experimentally, the force is found to vary proportionally to the square of the displacement, i.e. F(x)=−hx2 where h is some constant.
If the potential energy at x=0 m is U0=0 J, determine the potential energy at x=1.5 m.
If the potential energy at x=0 m is U0=0 J, determine the potential energy at x=1.5 m.
- 1.13 J
- 2.25 J
- 4.50 J
- 6.75 J
Q. A particle is released from rest at origin. It moves under the influence of a potential field, U=x2−3x. Find the Kinetic energy of the particle at x=2.
- 2 J
- 1 J
- 1.5 J
- 0 J
Q. Potential energy for a conservative force →F is given by U(x, y)=cos(x+y). Force acting on a particle at position given by coordinates (0, π4) is
- −1√2(^i+^j)
- 1√2(^i+^j)
- [12^i+√32^j]
- [12^i−√32^j]
Q. A particle is moving in a circular path of radius a with constant speed under the action of an attractive conservative force. Potential energy of the particle is given by the relation U=K2r2, where r is the radial distance of the particle from the centre of the circular path. Its total energy will be:
- K2a2
- Zero
- −32Ka2
- −K4a2
Q. The force exerted by a stretched chord at given displacements is shown in the table given below. Experimentally, the force is found to vary proportionally to the square of the displacement, i.e. F(x)=−hx2 where h is some constant.
If the potential energy at x=0 m is U0=0 J, determine the potential energy at x=1.5 m.
If the potential energy at x=0 m is U0=0 J, determine the potential energy at x=1.5 m.
- 1.13 J
- 2.25 J
- 4.50 J
- 6.75 J
Q. The potential energy of a particle of mass 5kg moving in the x−y plane is given by U=(−7x+24y)J, x and y being in a meter. If the particle starts from an origin, then the speed of the particle at t=2s is:
- 5m/s
- 14m/s
- 10m/s
- 17.5m/s