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Work & Energy Principles Lecture IX. Introduction In previous lecture, Newton ’ s 2 nd law (  F = ma) was applied to various problems of particle motion.

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Presentation on theme: "Work & Energy Principles Lecture IX. Introduction In previous lecture, Newton ’ s 2 nd law (  F = ma) was applied to various problems of particle motion."— Presentation transcript:

1 Work & Energy Principles Lecture IX

2 Introduction In previous lecture, Newton ’ s 2 nd law (  F = 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.

3 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 cos  where,  = the angle between the applied force and the displacement, ds = the magnitude of the displacement dr, F cos  = the tangential component of the force, F t. 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.

4 First: Work & Kinetic Energy – Cont. Work – Cont. Work is positive if F t is in the direction of the displacement; and work is negative if F t 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

5 First: Work & Kinetic Energy – Cont. Work – Kinetic Energy 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

6 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 ( e m ) Other sources of energy loss cause an overall efficiency of, e m < 1 Where, e e and e t is the electrical and thermal efficiencies, respectively.

7 Second: Potential Energy Gravitational Potential Energy V g Elastic Potential Energy V e or V g is positive, but  V g may be +ve or – ve. or V e is positive, but  V e may be +ve or – ve.

8 General Work and Energy Equation 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 Note: U ’ 1-2 is the work of all external forces other than gravitational and spring forces.

9 Work and Energy Principles Exercises

10 Exercise # 1 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 U f done on the crate by friction during the motion from A to B.

11 Exercise # 2 It takes 16 s to raise a 1270-kg car and the supporting 295-kg hydraulic car-lift platform to a height of 2 m. Knowing that the overall conversion efficiency from electric to mechanical power for the system is 82 percent, determine (a) the average output power delivered by the hydraulic pump to lift the system, (b) the average electric power required.

12 Exercise #3 A spring is used to stop a 90.7- kg package which is moving down a 20° incline. The spring has a constant k = 22 kN/m and is held by cables so that it is initially compressed 15 cm. Knowing that the velocity of the package is 2.4 m/s when it is 7.6 m from the spring and neglecting friction, determine the maximum additional deformation of the spring in bringing the package to rest.

13 Exercise # 4 A 4.5 kg collar is attached to a spring and slides without friction along a fixed rod in a vertical plane. The spring has an undeformed length of 35.6 cm. and a constant k = 700 N/m. Knowing that the collar is released from rest in the position shown, determine the speed of the collar at (a) point A, (b) point B.

14 Exercise # 5 An 4-kg collar A can slide without friction along a vertical rod and is released from rest in the position shown with the springs undeformed. Knowing that the constant of each spring is 300 N/m, determine the velocity of the collar after it has moved (a) 100 mm, (b) 190 mm.

15 Exercise # 6 The collar of weight W (25 N)is released from rest at A and travels along the smooth guide. Determine the speed of the collar just before it strikes the stop at B. The spring has an unstretched length L (30 cm), h (25 cm) and spring constant k of (4 N/cm).

16 Exercise # 7 The bob of mass M (0.75 kg) of a pendulum is fired from rest at position A by a spring which has a stiffness k (6 kN/m) and is compressed a distance δ (125 mm). Determine the speed of the bob and the tension in the cord when the bob is at positions B and C. Point B is located on the path where the radius of curvature is still r (0.6 m), i.e., just before the cord becomes horizontal.

17 Exercise # 8 A 907-kg car starts from rest at point (1) and moves without friction down the track shown determine: a) The normal force (N) exerted by the track on the car at point (2), where the radius of curvature of the track is 6 m b) The minimum safe value of the radius of curvature at point ( 3) ( the normal force (N) exerted by the track equal zero)


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