Restoring Force
Trending Questions
Q. The gravitational field in a region is given by the equation →E=(5^i+12^j) N/kg. If a particle of mass 2 kg is moved from the origin to the point (12 m, 5 m) in this region, the change in the gravitational potential energy is
- −480 J
- 480 J
- −240 J
- 240 J
Q. The variation of potential energy (U) with position (x) of a system is shown in the figure. The conservative force acting on the system is best represented by
Q. A body of mass 2 kg is projected from origin with a velocity 3 m/s along positive x− axis on which a force −x2 is acting. Find the maximum distance from origin upto which the body can go. (x is the position of particle)
- 6 m
- 5 m
- 3 m
- 4 m
Q. A particle of mass 10−2 kg is moving along the positive x-axis under the influence of a force F(x)=−α8x2, where α=10−2 N.m2. At time t=0, it is at x=1.0 m and velocity is v=0. Find the velocity when it reaches x=0.5 m. Consider the magnitude along with sign to represent the velocity.
- −0.5 m/s
- −1 m/s
- 1 m/s
- 2.5 m/s
Q. A particle of mass 10−2 kg is moving along the positive x-axis under the influence of a force F(x)=−K2x2 where K=10−2 Nm2. At time t=0, particle is at x=1 m and its velocity is v=0. The time at which it reaches x=0.25 m is
- 1.7 s
- 1.2 s
- 1.9 s
- 1.5 s
Q. A projectile starts from the origin at time t=0 and moves in the xy-plane with a constant acceleration α in the y-direction and constant speed along x direction. If equation of motion is y=βx2 then its velocity component in the x-direction will be
- 2αβ
- √2αβ
- α2β
- √α2β
Q. A particle which is constrained to move along x−axis is subjected to a force in the same direction which varies with distance x of the particle from the origin as F(x)=−kx+ax3. Here, k and a are positive constant. For x≥0, the functional form of the potential energy U(X) of the particle is
Q. A particle of mass 10−2 kg is moving along the positive x-axis under the influence of a force F(x)=−α8x2, where α=10−2 N.m2. At time t=0, it is at x=1.0 m and velocity is v=0. Find the velocity when it reaches x=0.5 m. Consider the magnitude along with the sign to represent the velocity.
- −0.5 m/s
- −1 m/s
- 1 m/s
- 2.5 m/s
Q. At time t=0 a particle starts moving along the x axis. If its kinetic energy increases uniformly with t, the net force acting on it must be
- Constant
- Proportional to t
- Inversely proportional to t2
- Proportional to 1/√t
Q.
The potential energy for a force field ¯F is given by ∪(x, y) = cos(x+y). The force acting on a particle at position given by coordinates (0, π/4) is