Question 1:
moderateA 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
Question 2:
moderatePotential energy of a particle at position x is given by U = x² – 5x. Which of the following is equilibrium position of the particle?
Question 3:
moderateIn 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
Question 4:
moderateIn 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
Question 5:
moderateA 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
Question 6:
moderateA 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
Question 7:
moderateThe 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
Question 8:
moderateA particle is placed at the origin and a force F = kx is acting on it (where k is positive constant). If U(0) = 0, the graph of U(x) versus x will be (where U is the potential energy function) :
Question 9:
moderateGiven below are two statements:
Assertion (A): A block of mass \(m\) is dropped from a height \(h\) on to a spring of spring constant \(k\). The maximum compression \(x_{\text{max}}\) in the spring is given by \(x_{\text{max}} = \sqrt{\frac{2mgh}{k}}\)
Reason (R): The work done by gravitational force is equal to the elastic potential energy stored in the spring at maximum compression.
In the light of the above statements, choose the correct answer from the options given below.
By conservation of energy, the total loss in gravitational potential energy is \(mg(h + x_{\text{max}})\), which equals the elastic potential energy stored in the spring \(\frac{1}{2}kx_{\text{max}}^2\). Thus, the given formula for \(x_{\text{max}}\) is incorrect as it neglects the term \(mgx_{\text{max}}\). Hence, (A) is false but (R) is true.
Question 10:
moderateThe potential energy of particle in a force field is \(U = \frac{A}{r} – \frac{B}{r^2}\) where \(A\) and \(B\) are positive constants and \(r\) is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is:
(2012 Pre)
For equilibrium, the force is zero: \(F = -\frac{dU}{dr} = 0\). Given \(U = \frac{A}{r} - \frac{B}{r^2}\), if the question implicitly means \(U = \frac{A}{r^2} - \frac{B}{r}\), then \(\frac{dU}{dr} = -\frac{2A}{r^3} + \frac{B}{r^2}\). Setting \(\frac{dU}{dr} = 0\) gives \(\frac{2A}{r^3} = \frac{B}{r^2} \Rightarrow r = \frac{2A}{B}\). Checking stability, \(\frac{d^2U}{dr^2} = \frac{6A}{r^4} - \frac{2B}{r^3}\), which is positive at \(r = \frac{2A}{B}\).