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.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/853383/original_4.png)
Which of the following graphs shows the correct variation of force F(x) with x?
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/853383/original_4.png)
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.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/869278/original_original_4a.png)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/869278/original_original_4a.png)
- 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