Lecture 2: Energy PDF

Lecture 2 Energy Introduction to Energy What is Energy ? The more energy you have, the more work you do  Introduction to Energy Energy, in physics, the capacity for doing work. work It may exist in potential, kinetic, thermal, electrical, chemical, nuclear...

Lecture 2 Energy Introduction to Energy What is Energy ? The more energy you have, the more work you do  Introduction to Energy Energy, in physics, the capacity for doing work. work It may exist in potential, kinetic, thermal, electrical, chemical, nuclear, or other various forms.  Introduction to Energy What is Work ? Work, in physics, a measure of energy transfer/change Watch that ! I will give some Energy to one of you, which will be transferred to Work  Work, math formula You gave this object an energy which transferred to Work, W = F. r The force F is constant The force F does work on the object in the direction of displacement F and r are Vectors  Work, math formula The object undergoes a displacement along a straight line while acted on by a constant force of magnitude F that makes an angle θ with the direction of the displacement. The work W done on a system is the product of the magnitude of the force F, the magnitude of the displacement Δr, and cos θ, where θ is the angle between the force and displacement vectors: Units of Work Work is a scalar quantity The unit of work is a joule (J) 1 joule = 1 newton. 1 meter = kg.m2/s2 J=N·m  Problem Which man is consuming less energy? Knowing that the displacement = 3 m Solution  Scalar Product of Two Vectors The scalar product of two vectors is written as A. B It is also called the dot product A. B = A B cos  is the angle between A and B The scalar product of any two vectors and is defined as a scalar quantity equal to the product of the magnitudes of the two vectors and the cosine of the angle  between them.  Scalar Product, cont The scalar product is commutative A.B = B.A The scalar product obeys the distributive law of multiplication A. (B + C) = A. B + A. C  What is the Work done by other forces? W = F.r = F r cos  The normal force, n, and the gravitational force, m g, do no work on the object cos  = cos 90° = 0 The force F does work on the object Concept of Differentiation and Integration Piece of unknown then integrate  Work Done by a Varying Force Assume that during a very small displacement, x, F is constant For that displacement, W ~ F x For all of the intervals, xf W  Fx x xi Work Done by a Varying Force, cont xf xf lim x 0  F x  xi x xi Fx dx xf Therefore,W x Fx dx i The work done is equal to the area under the curve  Problem: Calculate the work done according to the graph A force acting on a particle varies with x as shown in the Figure below. Calculate the work done by the force on the particle as it moves from x = 0 to x = 6 m. Solution Summary: Work  Hooke’s Law In the 1600s, a scientist called Robert Hooke discovered a law for elastic materials. He first stated the law in 1676. The law is named after 17th-century British physicist Robert Hooke. Hooke's equation holds in many other situations where an elastic body is deformed,  Hooke’s Law If a material returns to its original size and shape when you remove the forces stretching or deforming it (reversible deformation), we say that the material is demonstrating elastic behaviour. Inelastic material is one that stays deformed after you have taken the force away. If deformation remains (irreversible deformation) after the forces are removed then it is a sign of plastic behaviour. If you apply too big a force a material will lose its elasticity. Hooke discovered that the amount a spring stretches is proportional to the amount of force applied to it. This means if you double the force its extension will double, if you triple the force the extension will triple and so on. The elastic limit can be seen on the graph. This is where it stops obeying Hookes law. You can write Hooke's law as an equation: F = k∆x Where: F is the applied force (in newton, N), x is the extension (in metres, m) and k is the spring constant (in N/m). The extension ∆x can be found from: ∆x = stretched length – original length.  Hooke’s Law The force exerted by the spring is F = ±k ∆ x x is the position of the block with respect to the equilibrium position (x = 0) k is called the spring constant or elasticity constant and measures the stiffness of the spring The extension ∆x can be found from: ∆x = stretched length – original length.  Hooke’s Law, cont. When x is positive (spring is stretched), F is negative When x is 0 (at the equilibrium position), F is 0 When x is negative (spring is compressed), F is positive  Hooke's Law is often written: Fs = -kx This is because it also describes the force that the spring itself _______________ object exerts on an ___________ that is attached to it. The negative sign indicates that opposite the direction of the spring force is always _____________ to the displacement of the object -x compressed > 0 Fs ___ spring: Fs x=0 equilibrium ______________ undisturbed position, Fs = __ 0 spring +x stretched spring: < 0 Fs ___ Fs  Proble m A weight of 8 N is w/ weight w/o weight attached to a spring that has a spring constant of 160 N/m. How x much will the spring stretch? 8N  Solution Fs = 8N w/o weight w/ weight k= 160 N/m x =? x Fs = kx 8N 8 N = (160 N/m) x x = 5 x 10-2 m = 0.05 m  Hooke’s Law, final The force exerted by the spring is always directed opposite to the displacement from equilibrium F called the restoring force If the block is released it will oscillate back and forth between –x and x Calculate the Work Done by the Spring ? W=F.X  Work Done by a Spring Identify the block as the system Calculate the work as the block moves from xi = - xmax to xf = 0  Problem If a spring is stretched 2 cm by a suspended object having a mass of 0.55 kg, (a) what is the force constant of the spring? (b) How much work is done by the spring on the object as it stretches through this distance? (c) Evaluate the work done by the gravitational force on the object.  Solution Because the object is in equilibrium, the net force on it is zero and the upward spring force balances the downward gravitational force. (A) F = mg kd = mg , k = mg/d = (0.55)(9.8)/2x10-2 = 2.7x102 N/m  This work is negative because the spring force acts upward on the object, but its point of application moves downward.  Continue (c) If you expected the work done by gravity simply to be that done by the spring with a positive sign, you may be surprised by this result! To understand why that is not the case, we need to explore further, as we do in the Kinetic Energy and the Work–Kinetic Energy