* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download 12. Work Power & Energy
Elementary particle wikipedia , lookup
Hooke's law wikipedia , lookup
Brownian motion wikipedia , lookup
Virtual work wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Relativistic mechanics wikipedia , lookup
Matter wave wikipedia , lookup
Hunting oscillation wikipedia , lookup
Classical mechanics wikipedia , lookup
Electromagnetism wikipedia , lookup
Nuclear force wikipedia , lookup
Fictitious force wikipedia , lookup
Fundamental interaction wikipedia , lookup
Centrifugal force wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Centripetal force wikipedia , lookup
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 Work done by constant Force F F F F F cos F cos F cos F cos Displacement S Component of force in the direction of displacement = Fcos Work done = (Fcos) S =FS cos W = F.S Work by Varying Force Fx Fx Total work done is sum of all terms from xi to xf xf x xi x xf W = Fx Δx xi Work by Varying Force Fx xf W lim Fx x x 0 x i xf x xi xf W Fxdx xi Conservative Forces Work done on a particle between any two points is independent of the path taken by the particle. Force of gravity & spring Force Non-Conservative Forces Work done on a particle between any two points depends on the path taken by the particle. Frictional Force & Viscous force Illustration of principle Conservative Forces : Work done will be same Non-Conservative Force : Work done will be different Kinetic Energy Energy associated with the motion of a body. 1 KE = mv2 2 Nature : scalar Unit : joules(J) Work Energy Theorem Total work done by all the forces acting on a body is equal to the change in its kinetic energy. Wtotal = KEf - KEi Work done by Conservative forces (Wc) Non-Conservative forces (Wnc) External forces (Wext) Wc + Wnc + Wext = KEf - KEi Conservative Forces & Potential Energy Work done by a conservative force equals the decrease in the potential energy. xf Wc = Fxdx = -ΔU = Ui - Uf xi xf Uf = - Fxdx + Ui xi Ui can be assigned any value as only U is important Conservative Force: Spring force Force varies with position. Fs=-kx [k :Force constant] Fs 0 Fs kx Fs kx x x 0 x Work done in compression/extension of a spring by x 1 W = kx2 = PE stored in the spring 2 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 Solution : A specific force is external in origin and so is independent of reference frame. Hence answer is (d). Class Exercise - 2 A particle moves from a point r 1 2 i 3 j to another point r2 3 i 2 j during which a F 5 i 5 j 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 : Displacement d r2 – r1 i – j Fd 5 i j i – j 0 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 1 (a) zero (b) a bd d (c) 2a + bd 2 (d) (a + 2bd)d Solution : Force is variable. d bx 2 bd 2 W Fdx (a bx)dx ax ad 2 2 0 0 0 d d 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 Solution : Friction always acts opposite to the displacement. 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 Solution : Statement characterizes kinetic energy. 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) mgvtcos2 (c) mgvtsin2 (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 sin2 f vt mg Sin mg Class Exercise - 7 A force F –K y i x j (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) –2Ka2 (b) 2Ka2 (c) –Ka2 (d) Ka2 Solution First path: y = 0 F x j (perp. to displacement) W 0 Second path: x = a F ( K) y i a j y Displacement along x = 0 a 2 W Fydy –Ka 0 2 O 1 a a x Class Exercise - 8 Under the action of a force, a 2 kg body moves such that its position t3 x as a function of time is given by x 3 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 t3 dx d2 x 2 x , t , Fm 2mt 2 3 dt dt x 2 2 0 0 0 4 mt W Fdx 2mt t 2dt 2m t3dt 16 J 2 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: (a) Rg (b) (c) 2Rg (d) 2Rg Rg Solution PE = mgR R KE = 1 mv 2 2 PE = mgR KE 1 mv 2 2 1 mv 2 mgR v 2Rg 2 Hence answer is (b) 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. 4 cos37 5 , g 10 m/s (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. Mg sin37° = m N \ Work done by gravity = Work done against friction = Mg sin37° x s cos37 4 3 , sin37° 5 5 W 0.25 10 3 1 5 = 1.5 J Hence answer is (c) N F=mN Mg sin37o Mg cos37o 37o mg Thank you