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Potential Energy
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This lesson discusses the concept of potential energy and its derivation. It suggests and derives the relationship between the potential energy and work.

Adarsh Raj
Teaching Physics for IIT JEE /NEET. 20 MCQ for NEET daily In my feed. Question Series On Each chapter. HC VERMA EXERCISE solving .

Unacademy user
I don't think you should be making videos for Engineering students when you are still pursuing (learning) the subjects. I didn't find these important. Like in this video itself you should mention Harvard Architecture Von Neumann Architecture.
  1. Adarsh Raj E.C.E Educator at Unacademy Loves teaching and debating

  2. Potential Energy:- Potential energy of any body is the energy possessed by the body by virtue of its position or the state of deformation. With every potential energy there is an associated conservative force. The potential energy is measured as the magnitude of work done against the associated conservative force

  3. For example: (i) To place an object at any point in gravitational field work is to be done against gravitational field force. The magnitude of this work done against the gravitational force gives the measure of gravitational potential energy of the body at that position, which is U = mgh. Here h is the height of object from the reference level (ii) The magnitude of work done against the spring force to compress it gives the measure of elastic potential energy, which is U-1/2 Kx2 (ii) A charged body in any electrostatic field will have electrostatic potential energy

  4. Potential Energy W=Fx cos F = mg , x = h = 0, cos 0-1 W = mgh = PE Since this man is lifting the package upward at a CONSTANT SPEED, the kinetic energy is NOT CHANGING Therefore the work that he does goes into what is called the ENERGY OF POSITION or POTENTIAL ENERGY

  5. Potential Energy The man shown lifts a 10 kg package 2 meters above the ground. What is the potential energy given to the package by the man? PE = mgh h PE = (10)(9.8)(2)=

  6. Significance of potential energy: position U = U(x) From equation ( K + U(x) = 1 /2 m v2 + 1/2Kx2), Differentiating w.r.t. x, we find d/dx(K + U] = 0 since total mechanical energy is conserved. Therefore, dx L2 i.e. mv.dv/dx + kx = 0 Again v--a, the instantaneous acceleration of the system (block) at the position x 2 dv dx

  7. Therefore, 2 dx 2 2 i.e. mv.dv/dx + kx = 0 dv Again v--a, the instantaneous acceleration of the system (block) at the position Therefore, ma + kx = 0 l.e ie, d/dx K-ma and d/dx [U(x)] =-kx Hence we can say that the negative of differentiation of U(x) with respect to x, gives force acting on the system ma =-kx l.e. F(x) =- du (x) dx

  8. The potential energy of a system of two particles is given by U(x) = a/X-b/x. Find the minimum potential energy of the system, where x is the distance of separation, a, b are constants. U(x)b dx -2a b When the particle is in equilibrium F=0 2a b 2 a 0

  9. the minimum potential energy of the system is obtained by putting 2 a XU(X) - b abb min = 2a 4a 2a 4a

  10. A body of mass m dropped from a certain height strikes a light vertical fixed spring of stiffness k.Find the height of its fall before touching the spring if the maximum compression of the spring is equal to 3 mg The change in potential energy between A and B =-mg (h + x) The change in K. E. between A and B = 0 The change in spring potential energy = (1 /2) kx2 b-mg (h + x) + 0 +-kx2 = 0 Solution 2 2mg 3 mg 2mg 3mg 2k