Physics

Session Work, Power and Energy - 1

Session Objectives

Session Objective 1.Work done by constant force 2.Work done by variable force 3.Kinetic Energy 4.Work-Energy Theorem 5.Conservative and non-conservative forces 6.Potential Energy and Total Mechanical Energy

Displacement S Component of force in the direction of displacement = Fcos Work done = (FcosS =FS cos Work done by constant Force

Total work done is sum of all terms from x i to x f Work by Varying Force

Conservative Forces Non-Conservative Forces Force of gravity & spring Force Frictional Force & Viscous force Work done on a particle between any two points is independent of the path taken by the particle. Work done on a particle between any two points depends on the path taken by the particle.

Conservative Forces : Work done will be same Non-Conservative Force : Work done will be different Illustration of principle

Nature : scalar Unit : joules(J) Energy associated with the motion of a body. Kinetic Energy

Total work done by all the forces acting on a body is equal to the change in its kinetic energy. Conservative forces (W c ) Non-Conservative forces (W nc ) External forces (W ext ) Work done by Work Energy Theorem

U i can be assigned any value as only U is important Conservative Forces & Potential Energy Work done by a conservative force equals the decrease in the potential energy.

Force varies with position. F s =-kx[k :Force constant] Work done in compression/extension of a spring by x = PE stored in the spring Conservative Force: Spring force

Class Test

Class Exercise - 1 Force acting on a body is (a) dependent on the reference frame (b) independent of reference frame (c) dependent on the magnitude of velocity (d) None of these A specific force is external in origin and so is independent of reference frame. Solution : Hence answer is (d).

Class Exercise - 2 A particle moves from a point to another point during which a certain force acts on it. The work done by the force on the particle during displacement is: (a) 20 J(b) 25 J(c) zero(d) 18 J Solution : Hence answer is (c)

Class Exercise - 3 A force F = (a + bx) acts on a particle in the x-direction where a and b are constants. The work done by this force during a displacement from x = 0 to x = d is (a)zero(b) (c) 2a + bd(d) (a + 2bd)d Force is variable. Solution : Hence answer is (b)

Class Exercise - 4 A block starts from a point A, goes along a curvilinear path on a rough surface and comes back to the same point A. The work done by friction during the motion is: (a) positive(b) negative (c) zero(d) any of these Friction always acts opposite to the displacement. Solution : Hence answer is (b).

Class Exercise - 5 The work done by all forces on a system equals the change in (a)total energy (b) kinetic energy (c) potential energy (d) None of these Statement characterizes kinetic energy. Solution : Hence answer is (b).

Class Exercise - 6 A small block of mass m is kept on a rough inclined plane of inclination  fixed in an elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be (a)zero(b) mgvtcos 2  (c) mgvtsin 2 (d) mgvtsin2

Solution f = mg sin as block does not slide. Displacement d = vt Angle between d and f = (90 – ) W = fd cos(90° – ) = mgvt sin 2   Sin  mg vt f

Class Exercise - 7 A force (where K is a positive constant) acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force on the particle is: (a)–2Ka 2 (b) 2Ka 2 (c) –Ka 2 (d) Ka 2

Solution First path: y = 0 Second path: x = a Displacement along x = 0 y x a O 1 2 a

Class Exercise - 8 Under the action of a force, a 2 kg body moves such that its position x as a function of time is given by where x is in metre and t in seconds. The work done by the force in the first two seconds is: (a) 1,600 J(b) 160 J(c) 16 J(d) 1.6 J

Solution Hence answer is (c)

Class Exercise - 9 There is a hemispherical bowl of radius R. A block of mass m slides from the rim of the bowl to the bottom. The velocity of the block at the bottom will be:

Solution Hence answer is (b) PE = mgR R KE = mv 2 1 2

Class Exercise - 10 A block of mass 250 g slides down an incline of inclination 37° with uniform speed. Find the work done against the friction as the block slides down through 1.0 m. (a) 15 J(b) 150 J(c) 1.5 J(d) 1500 J

Solution As the block slides with uniform speed, net force along the incline is zero. Hence answer is (c) Mg sin37° =  N  Work done by gravity = Work done against friction = Mg sin37° x s = 1.5 J Mg cos37 o Mg sin37 o mg N F=N 37 o

Thank you