7.4) Kinetic Energy andThe Work-Kinetic Energy Theorem

1. d m F vi vf • 7.4) Kinetic Energy andThe Work-Kinetic Energy Theorem • Figure (7.13) - a particle of mass m moving to the right under the action of a constant net force F. • Because the force is constant - from Newton’s second law, the particle moves with a constant acceleration a. • If the particle is displaced a distance d, the net work done by the total force F is : • The relationships below are valid when a particle undergoes constant acceleration : • where vi = the speed at t = 0 and vf = the speed at time t. (7.12) Figure (7.13)

2. Substituting these expressions into Equation (7.12) : • The quantity represents the energy associated with the motion of the particle = kinetic energy. • The net work done on a particle by a constant net force F acting on it equals the change in kinetic energy of the particle. • In general, the kinetic energy K of a particle of mass m moving with a speed v is defined as : • Kinetic energy is a scalar quantity and has the same units as work = Joule (J). • It is often convenient to write Equation (7.13) in the form : (7.13) (7.14) (7.15) Work-kinetic energy theorem • That is :

3. Work-kinetic energy theorem • Must include all of the forces that do work on the particle in the calculation of the net work done. • The speed of a particle increases if the net work done on it is positive - because the final kinetic energy is greater than the initial kinetic energy. • The speed of a particle decreases if the net work done is negative because the final kinetic energy is less than the initial kinetic energy. • Kinetic energy is the work a particle can do in coming to rest, or the amount of energy stored in the particle. • We derived the wrok-kinetic energy theorem under the assumption of a constant net force - but it also is valid when the force varies. • Suppose the net force acting on a particle in the x direction is Fx. • Apply Newton’s second law, Fx = max, and use Equation (7.8), the net work done :

4. If the resultant force varies with x, the acceleration and speeed also depend on x. • Acceleration is expressed as : • Substituting this expression for a into the above equation for W gives : • Situation Involving Kinetic Friction • Analyze the motion of an object sliding on a horizontal surface - to describe the kinetic energy lost because of friction. • Suppose a book moving on a horizontal surface is given an initial horizontal velocity vi and slides a distance d before reaching a final velocity vf (Figure (7.15)). The net work done on a particle by the net force acting on it is equal to the change in its kinetic energy of the particle (7.16)

5. d vi vf fk Figure (7.15) • The external force that causes the book to undergo an acceleration in the negative x direction is the force of kinetic friction fk acting to the left, opposite the motion. • The initial kinetic energy of the book is ½mvi2 , and its final kinetic energy is ½mvf2 (Apply the Newton’s second law). • Because the only force acting on the book in the x direction is the friction force, Newton’s second law gives -fk = max. • Multiplying both sides of this expression by d and using Eq. (2.12) in the form vxf2 - vxi2 = 2axd for motion under constant acceleration give -fkd = (max)d = ½mvxf2 - ½mvxi2 or : (7.17a)

6. The amount by which the force of kinetic friction changes the kinetic energy of the book is equal to -fkd. • Part of this lost kinetic energy goes into warming up the book, and the rest goes into warming up the surface over which the book slides. • The quantity -fkd is equal to the work done by kinetic friction on the book plus the work done by kinetic friction on the surface. • When friction (as welll as other forces) - acts on an object, the work-kinetic energy theorem reads : • Wother represents the sum of the amounts of work done on the object by forces other than kinetic friction. (7.17b)

7. n vf n vf fk F F d d mg mg Example (7.7) : A Block Pulled on a Frictionless Surface A 6.0-kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force of 12 N. Find the speed of the block after it has moved 3.0 m. (a) (b) Figure (7.16)

8. L Example (7.8) : A Block Pulled on a Rough Surface Find the final speed of the block described in Example (7.7) if the surface is not frictionless but instead has a coefficient of kinetic friction of 0.15. Example (7.9) : Does the Ramp Lessen the Work Required? A man wishes to load a refrigerator onto a truck using a ramp, as shown in Figure (7.17). He claims that less work would be required to load the truck if the length L of the ramp were increased. Is his statement valid? Example (7.10) : Useful Physics for Safer Driving A certain car traveling at an initial speed v slides a distance d to a halt after its brakes lock. Assuming that the car’s initial speed is instead 2v at the moment the brakes lock, estimate the distance it slides. Figure (7.17)

9. Example (7.11) : A Block-Spring System A block of mass 1.6 kg is attached to a horizontal spring that has a force constant of 1.0x103 N/m (Figure (7.10)). The spring is compressed 2.0 cm and is then released from rest. (a) Calculate the speed of the block as it passes through the equilibrium position x = 0 if the surface is frictionless. • 7.5) Power • The two cars - has different engines (8 cylinder powerplant and 4 cylinder engine), same mass. • Both cars climb a roadway up a hill, but the car with the 8 cylinder engine takes much less time to reach the top. • Both cars have done the same amount of work against gravity, but in different time periods. • Know (i) the work done by the vehicles, and (ii) the rate at which it is done. • The time rate of doing work = power.

10. If an external force is applied to an object (assume acts as a particle), and if the work done by this force in the time interval t is W, then the average power expended during this interval is defined as : • The work done on the object contributes to the increase in the energy of the object. • Therefore, a more general definition of power is the time rate of energy transfer. • We can define the instantaneous power P as the limiting value of the average power as t approaches zero : Average power Represent increment of work done by dW t  0

11. From Equation (7.2), letting the displacement be ecpressed as ds, that dW = F·ds. • Therefore, the instantaneous power can be written : • The SI unit of power = joules per second (J /s) = watt (W). • A unit of power in the British engineering system is the horsepower (hp) : • A unit of energy (or work) can now be defined in terms of the unit of power. • One kilowatt hour (kWh) = the energy converted or consumed in 1 h at the constant rate of 1 kW = 1 000 J /s. v =ds / dt (7.18) 1W = 1 J/s = 1 kg ·m2 / s3 1 hp = 746 W kWh = a unit of energy, not power 1kWh = (103W)(3600 s) = 3.60 x 106 J

12. Example (7.12) : Power Delivered by an Elevator Motor An elevator car has a mass of 1 000 kg and is carrying passengers having a combined mass of 800 kg. A constant frictional force of 4 000 N retards its motion upward, as shown in Figure (7.18a). (a) What must be the minimum power delivered by the motor to lift the elevator car at a constant speed of 3.00 m/s?