Work, Energy & Power PDF

STEM 005 GENERAL PHYSICS 2  WORK, ENERGY AND POWER Work Mechanical Energy  Kinetic Energy  Work-Kinetic Energy Theorem  Potential Energy  Conservative and Nonconservative Forces  Conservation of Mechanical Energy Power  WORK WORK It i...

STEM 005 GENERAL PHYSICS 2  WORK, ENERGY AND POWER Work Mechanical Energy  Kinetic Energy  Work-Kinetic Energy Theorem  Potential Energy  Conservative and Nonconservative Forces  Conservation of Mechanical Energy Power  WORK WORK It is defined to be the product of the magnitude of the displacement times the component of the force parallel to the displacement. Work is a scalar quantity – it has In equation form, we can write no direction, but only magnitude, which can be positive or W = F ‖d negative.  WORK The product of the magnitude of the In equation form, we can write displacement times the component of the force parallel to the displacement. W = F ‖d F ) θ F cos θ d F ‖ = F cos θ  WORK In equation form, we can write W = F ‖d F ‖ = F cos θ W = F ‖d W = ( F cos θ ) d W = Fd cos θ  WORK Let’s consider that the force and displacement are parallel to each other, therefore, θ = 0 UNIT of Work W = Fd cos θ W = Fd cos θ W = ( 1 N )( 1 m ) cos 0 W = 1 N·m F W=1J d 1 Joule = 1 Newton-meter 1 J = 1 N·m  WORK ZERO WORK The force and the displacement are PERPENDICULAR to each other. W = Fd cos θ W = ( 1 N )( 1 m ) cos 90 FP W = 0 N·m θ = 90 W=0J d  WORK ZERO WORK The force and the displacement are PERPENDICULAR to each other. W = Fd cos θ W = ( 1 N )( 1 m ) cos 90 d W = 0 N·m Fg θ = 90 W=0J  WORK ZERO WORK W = Fd cos θ There is an Applied Force but NO displacement. F W = Fd cos θ W = ( 1 N )( 0 m ) cos 0 W = 0 N·m d=0 W=0J  WORK SAMPLE PROBLEM 1 WORK DONE ON CRATE A person pulls a 50-kg crate 40 m along horizontal floor by a constant force FP = 100 N, which acts at a 37o angle. The floor is rough and exerts a friction force Ffr = 50 N. a. Determine the work done by each force acting on the crate. b. Determine the net work done on the crate. WORK SAMPLE PROBLEM 1  WORK SAMPLE PROBLEM 1 WORK DONE ON CRATE A person pulls a 50-kg crate 40 m along horizontal floor by a constant force FP = 100 N, which acts at a 37o angle. The floor is rough and exerts a friction force Ffr = 50 N. a. Determine the work done by each force acting on the crate. b. Determine the net work done on the crate. FN FP ) θ = 37˚ dx Ffr Fg  WORK SAMPLE PROBLEM 1 - A WORK DONE ON CRATE A person pulls a 50-kg crate 40 m along horizontal floor by a constant force FP = 100 N, which acts at a 37o angle. The floor is rough and exerts a friction force Ffr = 50 N. a. Determine the work done by each force acting on the crate. Work done by GRAVITATIONAL FORCE Work done by NORMAL FORCE Work done by APPLIED FORCE (PULL) Work done by FRICTIONAL FORCE  WORK SAMPLE PROBLEM 1 - A WORK DONE ON CRATE A person pulls a 50-kg crate 40 m along horizontal floor by a constant force FP = 100 N, which acts at a 37o angle. The floor is rough and exerts a friction force Ffr = 50 N. a. Determine the work done by each force acting on the crate. Given: Equation: m = 50 kg FP = 100 N Wg = Fgd cosθ = mgd cosθ Ffr = 50 N θ = 37˚ WN = FNd cosθ = mgd cosθ Required: Wg WP = FPd cosθ WN WP Wfr = Ffrd cosθ Wfr  WORK SAMPLE PROBLEM 1 - A WORK DONE ON CRATE A person pulls a 50-kg crate 40 m along horizontal floor by a constant force FP = 100 N, which acts at a 37o angle. The floor is rough and exerts a friction force Ffr = 50 N. a. Determine the work done by each force acting on the crate. Work done by Gravitational force (Fg) Solution: d Wg = mgd cosθ 𝐦 θ = 90˚ = (50 kg)( 𝟗. 𝟖 𝐬𝟐 )(40 m) cos 90˚ Fg = 0 N·m =0J  WORK SAMPLE PROBLEM 1 - A WORK DONE ON CRATE A person pulls a 50-kg crate 40 m along horizontal floor by a constant force FP = 100 N, which acts at a 37o angle. The floor is rough and exerts a friction force Ffr = 50 N. a. Determine the work done by each force acting on the crate. Work done by Normal force (FN) Solution: FN θ = 90˚ WN = mgd cosθ 𝐦 d = (50 kg)( 𝟗. 𝟖 𝐬𝟐 )(40 m) cos 90˚ = 0 N·m =0J  WORK SAMPLE PROBLEM 1 - A WORK DONE ON CRATE A person pulls a 50-kg crate 40 m along horizontal floor by a constant force FP = 100 N, which acts at a 37o angle. The floor is rough and exerts a friction force Ffr = 50 N. a. Determine the work done by each force acting on the crate. Work done by Applied force / Pull (Fp) Solution: FP WP = FPd cosθ ) θ= 37˚ = (100 N)(40 m) cos 37˚ d = 3, 200.00 N·m = 3, 200.00 J  WORK SAMPLE PROBLEM 1 - A WORK DONE ON CRATE A person pulls a 50-kg crate 40 m along horizontal floor by a constant force FP = 100 N, which acts at a 37o angle. The floor is rough and exerts a friction force Ffr = 50 N. a. Determine the work done by each force acting on the crate. Work done by Frictional force (Ffr) Solution: θ = 180˚ Wfr = Ffrd cosθ ) = (50 N)(40 m) cos 180˚ Ffr d = - 2000 N·m = - 2000.00 J  WORK SAMPLE PROBLEM 1 - B WORK DONE ON CRATE A person pulls a 50-kg crate 40 m along horizontal floor by a constant force FP = 100 N, which acts at a 37o angle. The floor is rough and exerts a friction force Ffr = 50 N. b. Determine the net work done on crate. Given: WN Wg = 0 J WP WN = 0 J WP = 3,200.00 J Wfr = - 2000.00 J Wfr Required: Wnet (net work done) Wg  WORK SAMPLE PROBLEM 1 - B WORK DONE ON CRATE A person pulls a 50-kg crate 40 m along horizontal floor by a constant force FP = 100 N, which acts at a 37o angle. The floor is rough and exerts a friction force Ffr = 50 N. b. Determine the net work done on crate. Given: Equation: Wg = 0 J Wnet = Wg + WN + WP + Wfr WN = 0 J Solution: WP = 3,200.00 J Wnet = Wg + WN + WP + Wfr Wfr = - 2000.00 J = 0 J + 0 J + 3,200.00 J – 2000.00 J = 1,200.00 J Required: Answer: Wnet (net work done) Wnet = 1,200.00 J  FRICTION The force that resists the moving of an object over another. KINETIC FRICTION It resists the motion of a moving object. Fk = µk FN d STATIC FRICTION Coefficient of friction (µ) It resists the initiation of motion. Low µ = low friction Fs = µs FN Large µ = high friction  WORK SAMPLE PROBLEM 2 WORK DONE BY GRAVITATIONAL FORCE A 50.0 kg boy wanted to get a coconut from the tree. He can do this in two ways: (a) he can climb three or (b) use a ladder of length L = 11.0 m and inclined 33° with the horizontal. The height of the tree is 6.0 m. a. Find the work done by the gravitational force on the boy by climbing the tree, and b. Find the work done by the gravitational force on the boy by using a ladder.  WORK SAMPLE PROBLEM 2 - A WORK DONE BY GRAVITATIONAL FORCE A 50.0 kg boy wanted to get a coconut from the tree. He can do this in two ways: (a) he can climb three or (b) use a ladder of length L = 11.0 m and inclined 33° with the horizontal. The height of the tree is 6.0 m. a. Find the work done by the gravitational force on the boy by climbing the tree. d ) θ = 180˚ Fg  WORK SAMPLE PROBLEM 2 - A WORK DONE BY GRAVITATIONAL FORCE A 50.0 kg boy wanted to get a coconut from the tree. He can do this in two ways: (a) he can climb three or (b) use a ladder of length L = 11.0 m and inclined 33° with the horizontal. The height of the tree is 6.0 m. a. Find the work done by the gravitational force on the boy by climbing the tree. Given: Solution: m = 50.0 kg (mass of the boy) Wg = Fgd cosθ d = 6.0 m = mgd cosθ g = 9.8 m/s2 = (50.0 kg)( 𝐦 𝟗. 𝟖 𝟐 )(6.0m) cos 180˚ 𝐬 θ = 180˚ = - 2,940 N·m Required: = - 2,940 J Wg (work done by the grav. force) Equation: Answer: Wg = Fgd cosθ - 2,940.00 J = mgd cosθ  WORK SAMPLE PROBLEM 2 - B WORK DONE BY GRAVITATIONAL FORCE A 50.0 kg boy wanted to get a coconut from the tree. He can do this in two ways: (a) he can climb three or (b) use a ladder of length L = 11.0 m and inclined 33° with the horizontal. The height of the tree is 6.0 m. b. Find the work done by the gravitational force on the boy by using the ladder. d ) θ = 33˚ θ = 90˚ Fg  WORK SAMPLE PROBLEM 2 - B WORK DONE BY GRAVITATIONAL FORCE A 50.0 kg boy wanted to get a coconut from the tree. He can do this in two ways: (a) he can climb three or (b) use a ladder of length L = 11.0 m and inclined 33° with the horizontal. The height of the tree is 6.0 m. a. Find the work done by the gravitational force on the boy by using the ladder. d θ = 123˚ Fg  WORK SAMPLE PROBLEM 2 - B WORK DONE BY GRAVITATIONAL FORCE A 50.0 kg boy wanted to get a coconut from the tree. He can do this in two ways: (a) he can climb three or (b) use a ladder of length L = 11.0 m and inclined 33° with the horizontal. The height of the tree is 6.0 m. b. Find the work done by the gravitational force on the boy by using the ladder. Given: Solution: m = 50.0 kg (mass of the boy) Wg = Fgd cosθ d = 11 m (length of the ladder) = mgd cosθ g = 9.8 m/s2 𝐦 θ = 123˚ = (50.0 kg)( 𝟗. 𝟖 𝐬𝟐 )(11.0m) cos 123˚ Required: = - 2,910.6 N·m = - 2,910.6 J Wg (work done by the grav. force) Equation: Answer: Wg = Fgd cosθ - 2,910.60 J = mgd cosθ