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Lecture VII 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 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, dU Ft ds Note: Work is a Work is defined by the displacement multiplied by the force component in the direction of that scalar not a vector. displacement. 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: U Ft ds or U Fx dx Fy dy x2 U12 Fdx x1 x1 x2 kxdx 1k x2 2 x12 2 First: Work & Kinetic Energy – Cont. Work – Kinetic Energy The kinetic energy T of the particle is defined as: Note: T is a scalar; 1 and it is always +ve T mv 2 regardless of the 2 direction of v. 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: 2 2 2 U1 2 F d r ma d r ma t ds 1 1 1 dv dv ds dv But at v at ds vdv dt ds dt ds v2 U12 mv dv v1 1 2 m v2 v1 2 2 U12 T2 T1 T or T1 U12 T2 First: Work & Kinetic Energy – Cont. Power Power is a measure of machine capacity; it is the time rate of doing work; i.e. dU dr P F P F v dt dt 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) em Poutput Pinput em < 1 Other sources of energy loss cause an overall efficiency of, e em ee et Where, ee and et is the electrical and thermal efficiencies, respectively. Second: Potential Energy Gravitational Potential Energy Vg Vg mgh or Vg Vg 2 Vg1 mg h2 h1 mg h Vg is positive, but Vg may be +ve or –ve. Elastic Potential Energy Ve 0 x 1 2 Ve kx dx kx 2 or 1 2 Ve Ve2 Ve1 k x2 x1 2 2 Ve is positive, but Ve may be +ve or –ve. General Work and Energy Equation Note: U’1-2 is U12 T Vg Ve T Vg Ve E the work of all external forces other than or gravitational T1 Vg1 Ve1 U12 T2 Vg 2 Ve2 and spring forces. 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: E 0 or E Constant Work and Energy Principles Exercises Exercise # 1 3/103: The spring is unstretched when x = 0. If the body moves from the initial position x1 = 100 mm to the final position x2 = 200 mm, (a) determine the work done by the spring on the body and (b) determine the work done on the body by its weight. Exercise # 2 3/105: The 30-kg crate slides down the curved path in the vertical plane. If the crate has a velocity of 1.2 m/s down the incline at A and a velocity of 8 m/s at B, compute the work Uf done on the crate by friction during the motion from A to B. Exercise # 3 3/109: In the design of a spring bumper for a 1500-kg car, it is desired to bring the car to a stop from a speed of 8 km/h in a distance equal to 150 mm of spring deformation. Specify the required stiffness k for each of the two springs behind the bumper. The springs are undeformed at the start of impact. Exercise # 4 3/122: A department-store escalator handles a steady load of 30 people per minute in elevating them from the first to the second floor through a vertical rise of 7 m. The average person has a mass of 65 kg. If the motor which drives the unit delivers 3 kW, calculate the mechanical efficiency em of the system . Exercise # 5 3/130: Each of the sliders A and B has a mass of 2 kg and moves with negligible friction in its respective guide, with y being in the vertical direction. A 20-N horizontal force is applied to the midpoint of the connecting link of negligible mass, and the assembly is released from rest with q = 0. Calculate the velocity vA with which A strikes the horizontal guide when q = 90°. Exercise # 6 3/131: The ball is released from position A with a velocity of 3 m/s and swings in a vertical plane. At the bottom position, the cord strikes the fixed bar at B, and the ball continues to swing in the dashed arc. Calculate the velocity vC of the ball as it passes position C. Exercise # 7 3/143: The 0.9-kg collar is released from rest at A and slides freely up the inclined rod, striking the stop at B with a velocity v. The spring of stiffness k = 24 N/m has an unstretched length of 375 mm. Calculate v. Exercise # 8 3/144: The 4-kg slider is released from rest at A and slides with negligible friction down the circular rod in the vertical plane. Determine (a) the velocity v of the slider as it reaches the bottom at B and (b) the maximum deformation x of the spring. Exercise # 9 3/158: If the system is released from rest, determine the speeds of both masses after B has moved 1 m. Neglect friction and the masses of the pulleys.

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