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Physics. Session. Work, Power and Energy - 2. Session Objectives. Session Objective. Gravitational Potential Energy Potential Energy of an extended Spring Conservation of Energy Force and Potential Energy Relationship Conditions for equilibrium.
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Session Work, Power and Energy - 2
Session Objective • Gravitational Potential Energy • Potential Energy of an extended Spring • Conservation of Energy • Force and Potential Energy Relationship • Conditions for equilibrium
U (H) = mgH (at height H w.r.t : earth) =-W U is a properly of earth + block At height h U(h) = mgh U=mgh KE=1/2 mv2 = mg(H-h) KE(h) = mv2 = mg (H-h) H h E(h) = U(h)+K(h) Gravitational Potential Energy
M M M X= 0 x xo Potential energy : spring Kinetic energy : mass Potential Energy of an Extended Spring At maximum extension (spring at rest)
E,U,K K U - x0 x0 x=0 Conservation of energy of isolated System System : spring + mass Spring energy E(x) = U(x) + K(x)
Conservation of energy of isolated System General Rule For an isolated system, in the absence of non – conservative forces , the total mechanical energy remains a constant
Conservation of Energy Three kinds of energy act on a system : • Mutual forces : internal and • conservative work = - PE 2. Mutual forces : internal and non conservative work changes mechanical energy 3. External forces : Work changes total energy
Equilibrium along x (1) Equilibrium along y (2) U Equilibrium along z (3) q x - q = 0 + Linear equilibrium Spring - mass system Condition of Equilibrium of a system under linear motion
Some useful relations Work done by pressure on a fluid :
Power Rate of doing work is power Power = Rate of Energy transfer Nature : scalar Units : Watt(W) 1 W = 1 J/s 1 hp = 746 W
A body of mass m was slowly pulled up the hill by a force F which at each point was directed along the tangent of the trajectory. All surfaces are smooth. Find the work performed by this force. • mgl (b) –mgl • (c) mgh (d) Zero l Class Exercise - 1
Solution Surfaces are smooth, so no friction exists. The force F is always tangential. Work is done against gravity is conservative. So work done is path independent and equal to increase in potential energy : W = mgh Hence answer is (c)
A particle is moving in a region where potential U is given by The force acting on the particle is Class Exercise - 2
Solution U = K(x2 + y2 + z2)
The potential energy of a particle in a conservative field has the form , where a and b are positive constants, r is the distance from the centre of the field. Then (a) at exists a stable equilibrium (b) at exists an unstable equilibrium (c) at exists a stable equilibrium (d) at exists an unstable equilibrium Class Exercise - 3
To check stability, So at there is a stable equilibrium. Solution For equilibrium: Putting the value of r, which is a position of stable equilibrium. Hence answer is (c).
If the KE of a particle is doubled, then its momentum will • remain unchanged • (b) be doubled • (c) be quadrupled • (d) increase by times Class Exercise - 6
Solution Hence answer is (d)
An engine pumps up 100 kg of water through a height of 10 m in 5 s. Given that the efficiency of engine is 60%, what is the power of the engine? • 33 kW (b) 3.3 kW • (c) 0.33 kW (d) 0.033 kW Class Exercise - 7
Solution Work done = 100 × 10 = 1000 J Hence answer is (c)
Class Exercise - 8 A system has two light springs with stiffness k1 and k2 joined in series and hanging from a rigid support. What is the minimum work that needs to be done to stretch the system by a lengthDl?
Solution At all time the tension T in the two springs is the same. The extensionDl = Dl1 + Dl2 … (i) If we assume both springs are replaced by a single spring of stiffness k: And the minimum work to be done on the system (equivalent spring) is
Find out whether the field of force is conservative, a being a positive constant Class Exercise - 9
acting is not conservative. Solution Consider a particle at origin. It is taken along the perimeter of unit square as shown in the figure. Total work done on the particle: W = W1 + W2 + W3+ W4 = 0 + a + 0 + 0 = a
A particle moves along the X-axis through a region in which the potential energy U(x) varies as U(x) = 4x – x2. • Find the position of the particle when • the force on it is zero (ii) The particle has a constant mechanical energy of 4.0 J. Find the kinetic energy as a function of x. Class Exercise - 10
Solution F = 0 Þ 4 – 2x = 0 or x = 2m b. Total energy = K(x) + U(x) = 4 J Þ K(x) + 4x – x2 = 4 Þ K(x) = x2 – 4x + 4