Work & Energy Principles Prepared by Dr. Hassan Fadag. Lecture IX Work & Energy Principles

Introduction In previous lecture, Newton’s 2nd law (SF = ma) was applied to various problems of particle motion to establish the instantaneous relationship between the net force acting on a particle and the resulting acceleration of this particle. To get the velocity and displacement, the appropriate kinematics equations may be applied. There are two general classes of problems in which the cumulative effects of the unbalanced forces acting on a particle are of interest: 1) Integration of the forces w.r.t. the displacement of the particle. This leads to the equations of work and energy. 2) Integration of the forces w.r.t. the time. This leads to the equations of impulse and momentum. Incorporation of the results of these integrations directly into the governing equations of motion makes it unnecessary to solve directly for the acceleration.

First: Work & Kinetic Energy Work done by the force F during the displacement dr is defined by: dU = F . dr The magnitude of this dot product is: dU = F ds cosa where, a= the angle between the applied force and the displacement, ds = the magnitude of the displacement dr, F cosa = the tangential component of the force, Ft. Thus, Work is defined by the displacement multiplied by the force component in the direction of that displacement. Note: Work is a scalar not a vector.

First: Work & Kinetic Energy – Cont. Work – Cont. Work is positive if Ft is in the direction of the displacement; and work is negative if Ft is in opposite direction to the displacement. Forces which do work are termed active forces, while constraint force which do no work are termed reactive forces. Work SI unit is Joule (J); where 1 J = 1 N.m During a finite movement of the point of application of a force, the force does an amount work equal to: or

First: Work & Kinetic Energy – Cont. The kinetic energy T of the particle is defined as: It is the total work that must be done on the particle to bring it from a state of rest to a velocity v. Work and Kinetic Energy relation: Note: T is a scalar; and it is always +ve regardless of the direction of v. or

First: Work & Kinetic Energy – Cont. Power Power is a measure of machine capacity; it is the time rate of doing work; i.e. Power is scalar and its SI unit is watt (W), where 1 W = 1 J/s 1 hp = 550 ft-lb/sec = 33,000 ft-lb/min 1 hp = 746 W = 0.746 kW Mechanical Efficiency (em) Other sources of energy loss cause an overall efficiency of, em < 1 Where, ee and et is the electrical and thermal efficiencies, respectively.

Second: Potential Energy Gravitational Potential Energy Vg or Vg is positive, but DVg may be +ve or –ve. Elastic Potential Energy Ve or Ve is positive, but DVe may be +ve or –ve.

General Work and Energy Equation Note: U’1-2 is the work of all external forces other than gravitational and spring forces. or For problems where the only forces are the gravitational, elastic, and nonworking constraint forces, the U’1-2-term is zero, and the energy equation becomes: or

Work and Energy Principles Exercises

Exercise # 1 Calculate the velocity v of the 50-kg crate when it reaches the bottom of the chute at B if it is given an initial velocity of 4 m/s down the chute at A. The coefficient of kinetic friction is 0.30.

Exercise # 2 The crate, which has a mass of 100 kg, is subjected to the action of the two forces. If it is originally at rest, determine the distance it slides in order to attain a speed of 6 m/s. The coefficient of kinetic friction between the crate and the surface is μk = 0.2.

Exercise # 3 The skier starts from rest at A and travels down the ramp. If friction and air resistance can be neglected, determine his speed vB when he reaches B. Also, find the distance s to where he strikes the ground at C, if he makes the jump traveling horizontally at B. Neglect the skier’s size. He has a mass of 70 kg.

Exercise # 4 For a short time the crane lifts the 2.50-Mg beam with a force of F = (28 + 3s2) kN. Determine the speed of the beam when it has risen s = 3 m. How much time does it take to attain this height starting from rest.

Exercise # 5 The elevator E and its freight have a total mass of 400 kg. Hoisting is provided by the motor M and the 60-kg block C. If the motor has an efficiency of e = 0.6, determine the power that must be supplied to the motor when the elevator is hoisted upward at a constant speed of vE = 4 m/s.

Exercise # 6 An automobile having a mass of 2 Mg travels up a 7° slope at a constant speed of v = 100 km/h. If mechanical friction and wind resistance are neglected, determine the power developed by the engine if the automobile has an efficiency e = 0.65.

Exercise # 7 The 1.2-kg slider is released from rest in position A and slides without friction along the vertical-plane guide shown. Determine (a) the speed vB of the slider as it passes position B and (b) the maximum deflection d of the spring.

Exercise # 8 The gantry structure is used to test the response of an airplane during a clash. The plane of mass 8-Mg is hoisted back until θ = 60°, and then pull-back cable AC is released when the plane is at rest. Determine the speed of the plane just before clashing into the ground, θ = 15°. Also, what is the maximum tension developed in the supporting cable during the motion?

Exercise # 9 The 7-kg collar A slides with negligible friction on the fixed vertical shaft. When the collar is released from rest at the bottom position shown, it moves up the shaft under the action of the constant force F = 200 N applied to the cable. Calculate the stiffness k which the spring must have if its maximum compression is to be limited to 75 mm. The position of the small pulley at B is fixed.

Exercise # 10 Cylinder A has a mass of 3 kg and cylinder B has a mass of 8 kg. Determine the speed of A after it moves upwards 2 m starting from rest. Neglect the mass of the cord and pulleys.