1. Sect. 8-7: Gravitational Potential Energy & Escape Velocity

2. As we’ve already seen, far from the surface of the Earth, the force of gravity is not constant: From the formal definition of work we had earlier, the work done on an object moving in the Earth’s gravitational field is given by:

3. Integrating gives: Because the value of the integral depends only on the end points, the Gravitational Force is Conservative & we can define a gravitational potential energy:

4. Ex. 8-12: Package Dropped from a High Speed Rocket A box of empty film canisters is allowed to fall from a rocket traveling outward from Earth at a speed of 1800 m/s when 1600 km above the Earth’s surface. The package eventually falls to the Earth. Calculate its speed just before impact. (Ignore air resistance.)

5. If an object’s initial kinetic energy is equal to the potential energy at the Earth’s surface, its total energy will be zero. The velocity at which this is true is called the ESCAPE VELOCITY. For Earth this is:

6. Example 8-13: Escaping the Earth or the Moon a. Compare the escape velocities of a rocket from the Earth and from the Moon b. Compare the energies required to launch the rockets. Moon: M M = 7.35  10 22 kg, r M = 1.74  10 6 m Earth: M E = 5.98  10 24 kg, r E = 6.38  10 6 m

7. Sect. 8-8: Power Power  The rate at which work is done or the rate at which energy is transformed. Instantaneous power: Average Power: P = (Work)/(Time) = (Energy)/(time) SI units: Joule/Second = Watt (W) 1 W = 1J/s British units: Horsepower (hp). 1hp = 746 W “Kilowatt-Hours” (from your power bill). Energy! 1 KWh = (10 3 Watt)  (3600 s) = 3.6  10 6 W s = 3.6  10 6 J

8. Example 8-14: Stair Climbing Power A 60-kg jogger runs up a flight of stairs in 4.0 s. The vertical height of the stairs is 4.5 m.. a. Estimate the jogger’s power output in watts and horsepower. b. How much energy did this require?

9. Power is also needed for acceleration & for moving against the force of friction. The power can be written in terms of the net force & the velocity:

10. Average Power: P = W/t Its often convenient to write power in terms of force & speed. For example, for a force F & displacement d in the same direction, we know: W = F d  P = F (d/t) = F v = average power v  Average speed of object

11. Example: Power Delivered by an Elevator Motor An elevator car of mass m e = 1,600 kg, carries people, mass m p = 200 kg. A constant friction force f k = 4,000 N acts against the motion. (A) Find the power needed for a motor to lift the car + passengers at a constant velocity v = 3 m/s. (B) Find power needed for a motor to lift the car at instantaneous velocity v with an upward acceleration a = 1.0 m/s 2.

12. Example 8-15: Power needs of a car Calculate the power required of a 1400-kg car to do the following: a. Climb a 10° hill (steep!) at a steady 80 km/h b. Accelerate on a level road from 90 to 110 km/h in 6.0 s Assume that the average retarding force on the car is F R = 700 N. a. ∑F x = 0 F – F R – mgsinθ = 0 F = F R + mgsinθ P = Fv b. Now, θ = 0 ∑F x = ma F – F R = ma v = v 0 + at P = Fv