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This chapter explores the concepts of work and energy, including work done by both constant and varying forces, the work-kinetic energy theorem, nonisolated systems, kinetic friction, and power systems.
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Energy and energy transfer (chapter six) • Work done by a constant force • Scalar product • Work done by a varying force • Kinetic energy, Work-kinetic energy theorem • Nonisolated systems • Kinetic friction • Power
Systems In discussion of work and energy, it is important that we are clear about the objects we are considering. A system is an object, group of objects or region of space with well defined boundaries.
Work done by a constant force • Effect of force acting on an object over some distance. • The component of the force in the direction of the displacement, multiplied by the magnitude of the displacement (or the component of the displacement in the direction of the force times the magnitude of the force.) • Result is a scalar quantity • Units: N·mJ b(Joule)
Scalar products product of two vectors: where is the angle between the two vectors In terms of components:
Properties of scalar product • Commutative • Distributive for the unit vectors
Work done by a varying force • Increment of work: force applied over small displacement • Total work: sum of increments • Taking the limit for infinitesimally small displacements
Example: work done pushing a block up a frictionless inclined plane at constant velocity
Question • What do Winnie the Pooh and Attila the Hun have in common?
Question • What do Winnie the Pooh and Attila the Hun have in common? • Answer: Same middle name.
Kinetic energy A net force acting on an object in the x-direction will do an amount of work Using Newton’s 2nd law
Kinetic energy From this expression we define the kinetic energy and the expression for net work becomes This is the work-kinetic energy theorem
Example A block slides down a frictionless plane – what is the velocity at the bottom? Forces perpendicular to the plane do no work
Example – work done by a spring • Force exerted by spring (Hooke’s law) where x is the displacement of the spring from its unstretched position The work done by the spring is
Example – work done by a spring A spring (k=100 N/m) is slowly stretched by 2 cm from its unstretched length. What is the work done by the force of the spring? The result is negative since the force is applied in the opposite direction to the displacement. What would be the work done by the spring if it were compressed by the same amount? What is the work done by the force stretching or compressing the spring?
v Example – speed of a block on a spring The same spring in the last example is put in contact with a block (mass m=1.00 kg), compressed 2.00 cm and then let go. How fast is the block moving when it loses contact with the spring? Work done by the spring (now positive – spring is returning to unstretched length) Must equal the change in kinetic energy
Example –block dropped onto a spring Same spring, same mass, now the mass is dropped from a height 1.00 m above the uncompressed spring. How far down does the spring compress? Work done by gravity on the block is 1.00m + d The work done on the block by the spring is Since the change in kinetic energy is zero, the total work done must also be zero
Nonisolated systems • Work can be thought of as the transfer of energy between a system and its environment • Forms of energy of a system other than kinetic: internal (thermal) • Ways of energy transfer other than work (mechanical waves, heat, matter transfer, electrical transmission, EM radiation)
Kinetic friction The new form of the work-KE theorem: looks the same, but note that now the application point of the force (friction) is changing For the large system of both objects
Example A block slides down an inclined plane with coefficient of kinetic friction k– what is the velocity at the bottom?
Power • power = rate of energy transfer • average power • instantaneous power • more general definition: • Units: Watt (=J/s) hp=550 ft·lb = 746 W
Example How much power is needed to accelerate a 1.00103 kg car from 0-60.0 mph in 5.00 seconds, ignoring air resistance and friction? How much is needed to keep it moving at 60.0 mph if friction and air resistance equal 1.00 102 N? Assuming a constant acceleration (60 mph = 26.8 m/s)
To keep it moving at a constant velocity, the magnitude of the applied force must equal that from the air resistance and friction