Potential Energy
Trending Questions
Gravitational Potential is?
The SI unit of gravitational potential is
A rod of mass m and length l is lying on a horizontal table. The work done in making it stand vertically on one end will be
- x=−1/2 is position of stable equilibrium
- x=2 is position of stable equilibrium
- x=−1/2 is position of unstable equilibrium
- x=2 is position of neutral equilibrium
- V=E(R+r)
- V=E
- V=E(1+r/R)
- V=E(1+R/r)
Is gravitational potential a vector?
The potential energy U in joule of a particle of mass 1 kg moving the x-y plane obeys the law U = 3x + 4y, where (x, y) are the co-ordinates of the particle in meter. If the particle is at rest at (6, 4) at time t = 0 then:
The particle has constant acceleration
The particle has zero acceleration
The speed of the particle when it crosses y-axis is 10 m/s
Co-ordinate of particle at t = 1 sec is (4.5, 2)
- 1 s
- 3 s
- 4 s
- 2 s
The correct option is
- The time taken in travelling DAB is less than that for BCD
- The time taken in travelling DAB is greater than that for BCD
- The time taken in travelling CDA is less than that for ABC
- The time taken in travelling CDA is greater than that for ABC
- 8
- −8
- 4
- −4
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 metre. If the body is at rest at point (6 m, 4m) at time t = 0, it will cross the y-axis at time t equal to
1 s
2 s
3 s
4 s
- The maximum kinetic energy of the particle is 16 J
- Both (a) and (b) are correct
- On negative x-axis particle travels upto x=−2 m
- Both (a) and (b) are wrong
A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.
(a) How much work does she do against the gravitational force?
(b) Fat supplies 3.8×107 J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
An electron initially at rest is accelerated through a potential difference of . The energy gained by the electron is:
- U=−Gm2lln(2a+l2a−l)
- U=−Gm2lln(a+la)
- U=−2Gm2lln(2a+l2a−l)
- U=−Gm2lln(aa−l)
Once a zero potential energy reference state is chosen, the changes in potential energy for a given initial and final states
Are same
Are different
Depend strictly on the choice of the zero of potential energy
Become indeterminate
- True
- False
- 1:3
- 2:3
- 3:2
- 3:1
- 6Vm−1 along negative x-axis
- 6Vm−1 along positive x-axis
- 12Vm−1 along positive x-axis
- 12Vm−1 along negative x-axis
- 8Vm−1 along negative x-axis
U(y)=k[1−e−y2] for −a≤y≤a where k is a constant. Then
- At y=0, potential energy =0
- At y=±a, potential energy is maximum
- At y=0, kinetic energy is maximum
- At y=±a, Kinetic energy is minimum
- 1
- −1
- 2
- 16
Statement(A): Negative charges always move from a higher potential to lower potential point.
Statement (B): Electric potential is vector.
- A is true but B is false
- B is true but A is false
- Both A and B false
- Both A and R are true