1. 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 v i = the speed at t = 0 and v f = the speed at time t. (7.12) m d FF vivi vfvf 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  F x. Apply Newton’s second law,  F x = ma x, 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 v i and slides a distance d before reaching a final velocity v f (Figure (7.15)). (7.16) 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

5. fkfk vivi vfvf d 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 f k acting to the left, opposite the motion. The initial kinetic energy of the book is ½mv i 2, and its final kinetic energy is ½mv f 2 (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 -f k = ma x. Multiplying both sides of this expression by d and using Eq. (2.12) in the form v xf 2 - v xi 2 = 2a x d for motion under constant acceleration give -f k d = (ma x )d = ½mv xf 2 - ½mv xi 2 or : (7.17a)

6. The amount by which the force of kinetic friction changes the kinetic energy of the book is equal to -f k d. 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 -f k d 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 :  W other represents the sum of the amounts of work done on the object by forces other than kinetic friction. (7.17b)

7. 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. n vfvf F d mgmg (a) n vfvf F d mgmg fkfk (b) Figure (7.16)

8. 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. L 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 1.0x10 3 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  t  0 Represent increment of work done by dW

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. (7.18) v =ds / dt 1W = 1 J/s = 1 kg ·m 2 / s 3 1 hp = 746 W 1kWh = (10 3 W)(3600 s) = 3.60 x 10 6 J kWh = a unit of energy, not power

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?