Tiêu đề WORK, POWER and ENERGY.pdf ✅

CLASS VIII PHYSICS (i) If θ is acute, W is positive (force tries to increase the speed of the body). (ii) If θ = 90∘ i.e., force is perpendicular to displacement, then W = 0 i.e., Work done = W = mg.S. cos 90∘ = 0 iii) If θ is obtuse, " W is negative (force tries to decrease the speed of the body) iv) If F⃗ is in the direction of S⃗, then W = FS(θ = 0 ∘ ) v) If F⃗ is opposite to S⃗, then W = −FS(θ = 180∘ ) Applications: a) If we lift a body from rest to a height h

CLASS VIII PHYSICS 3. Net force acting on the body is zero. As W = ∫ F⃗ ⋅ ds⃗ so, if ds⃗ = 0, W = 0 i.e., if the displacement of a particle or body is zero whatever be the force, work done is zero (except non-conservative force). a) When a person tries to displace a wall or stone by applying a force (actually its centre of mass does not change) the work done is zero. b) A weight lifter does work in lifting the weight from the ground but does not work in holding it up. As W = ∫ Fdscos θ, so W = 0, if θ = 90∘ , i.e., if force is always perpendicular to motion, work done by the force will be zero though neither force nor displacement is zero. This is why: a) When a porter moves with a suitcase on his head on a horizontal level road, the work done by the lifting force or force of gravity is zero. b) When a body moves in a circle the work done by the centripetal force is always zero. c) When the bob of a simple pendulum swings, the work done by tension in the string is zero. Applications on work: 1. If a force is changing linearly from F1 to F2 over a displacement S then work done is W = ( F1+F2 2 ) S If a force displaces the particle from its initial position r⃗i to final position r⃗f then displacement vector is S⃗ = r⃗f − r⃗i . W = F⃗ ⋅ S⃗ = F⃗ ⋅ (r⃗f⃗ − ri ⃗⃗) 2. Work done in pulling the bob of mass m of a simple pendulum of length L through an angle θ to vertical by means of a horizontal force F.