Conservative Force as Gradient of Potential
Trending Questions
Q. The potential energy of a body of mass 2 kg in a conservative field is U=(6x−8y) J. If the initial velocity of the body is →u=(−1.5^i+2^j) m/s, then the total distance travelled by the particle in the first two seconds is
- 10 m
- 12 m
- 15 m
- 18 m
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
- −8.39 N
- 14.8 N
- −14.8 N
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. 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 potential energy of 1 kg particle moving 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. A particle is moving in a circular path of radius 'a' under the action of an attractive potential, U=−k2r2 (where r is the radial distance). Its total energy is
- Zero
- −32ka2
- −k4a2
- k2a2
Q. The potential energy (in joule) of a body of mass 2 kg moving in the x−y plane is given by U=6x+8y, where the position coordinates x and y are measured in metres. The body is at rest at point (6 m, 4 m) at time t=0. It will cross the y− axis at time t equal to
- 1 s
- 2 s
- 3 s
- 4 s
Q. The potential energy (in joule) of a body of mass 2 kg moving in the x−y plane is given by U=6x+8y, where the position coordinates x and y are measured in metres. The body is at rest at point (6 m, 4 m) at time t=0. It will cross the y− axis at time t equal to
- 1 s
- 2 s
- 3 s
- 4 s
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 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. 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
- −8.39 N
- 14.8 N
- −14.8 N
Q. The potential energy U in joules of a particle of mass 1 kg moving in the x−y plane obeys the law U=3x+4y, where (x, y) are the co-ordinates of the particle in metres. If the particle is released from rest at (6, 4) at time t=0, then
- the particle has zero acceleration
- co-ordinates of the particle at t=1 s is (2, 4.5)
- co-ordinates of the particle at t=1 s is (4.5, 2)
- co-ordinates of the particle at t=1 s is (2, 2)
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 varying 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. The potential energy function U(x) of a particle moving along x direction is given by U(x)=ax2−bx. Find the equilibirum point (xe).
- xe=a2b
- xe=b2a
- xe=a2b2
- xe=2ab
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+z^k
- −x^i−y^j
- x^i−y^j
Q. The potential energy of a body of mass 2 kg in a conservative field is U=(6x−8y) J. If the initial velocity of the body is →u=(−1.5^i+2^j) m/s, then the total distance travelled by the particle in the first two seconds is
- 10 m
- 12 m
- 15 m
- 18 m
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