A cubical vessel of height 1 m is full of water. The minimum work done in taking water-out from vessel will be
As side length of cubical tank is 1 m. Volume is 1 m³. Volume of water in 1 m³ is 1000 lit and mass will be 1000 kg.
Height of center of mass of the tank will be 1/2 m.
Potential Energy = 1000 ×10 × 1/2 = 5000 J
A man pulls a bucket full of water from h metre deep well. If the mass of rope is m and mass of bucket full of water is M, then work done by the man is
Bucket of mass M will rise by a distance of h.
While center of mass of the rope rises by height of h/2.
So, Work Done = (M+m/2).g.h
Potential energy of a particle at position x is given by U = x² – 5x. Which of the following is equilibrium position of the particle?
In a conservative field, the potential energy U as a function of position x is given by U = x². Then the corresponding conservative force is given by
Select correct statement regarding stable equilibrium
All the statements are true regarding stable equilibrium
In a certain field, the potential energy is U = ax² – bx³, where a and b constants. The particle is in stable equilibrium at x equal to
Potential Energy is U = ax² – bx³. For Equilibrium F = -dU/dx
⇒ dU/dx= 2ax- 3bx²=0
⇒ x = 2a/3b
A body of mass m is raised to height h from ground along three different paths viz I, II and III against gravity as shown below. If WI, WII and WIII are the work done along the respective paths of I, II and III, then the correct option is (there is no nonconservative force)
Work done by conservative force is independent of path taken so, WI = WII = WIII
A uniform chain has a mass m and length l. It is held on a frictionless table with one-sixth of its length hanging over the edge. The work done in just pulling the hanging part back on the table is
mass of hanging portion = m/6
Potential energy of handing portion = m/6×g×(-l/12)= -mgl/72
Work done by external agent is negative of change is Potential Energy = mgl/72
The potential energy of a particle oscillating on x-axis is given as : U = 20 + (x – 2)2 Joules. If total mechanical energy of the particle is 36 J, then its maximum kinetic energy is:
Given U = 20 + (x – 2)2
For Equilibrium , dU/dx = 2 (x-2) =0, So, x=2 is the point of Equilibrium.
d²U/dx²= 2 >0 so, x= 2 is the point of stable equilibrium.
U min = 20 J
As Total Energy is 36 J.
U min + K max = 36 J
⇒ K max = 16 J
If potential energy between electron and proton at a distance r is given by U =-(ke²/3r³), then force acting is
Force F = - dU/dr = d ((ke²/3r³))/dr = ke²/r^4