1. Physics 211 Work done by a constant force Work done by a varying force Kinetic energy and the Work-Energy theorem Power 6: Work and Energy

2. If a total non zero force acts on an object it effects the state of the object The longer it acts the more it effects the object If the object is initially at rest it can cause the object to move If the force keeps acting on the object then it acts while the object is at different positions

3. A measure of how much it effects the object is the distance over which the force acts on the object We describe this effect by the quality WORK We quantify it by the amount of work done by the force The force does work on the object

4. The work done by an agent exerting a constant force is the product of the force in the direction of displacement times the magnitude of the displacement F d

5. If the force is not pointing in the direction of displacement then it is ONLY the component of that force in the direction of displacement that ”does” the work F  d

6. recall 3 i  2j  4k   2 i  5j  2k   3  2  2  (  5)  (  4)  2  12  3 i  2j  4k 2 i  5j  2k cos  SI units of work W   F  d   Nm  Joules( J)

7. Work done by a varying force For an object moving alongx- axis W  W i   F x i  d i  Fx i    x i In order that portions of force are constant over the portions of the displacement  x i W  lim  x i   0 F x x i    x i       F x x  dx x initial x final  = area under the curve of the function F x x  between x initial and x final

8. Force exerted by spring on a mass m F(x)  kx where x is the displacement from the equilibrium length, and k is the force constant of the spring ( x  0 if spring is extended; x  0 if spring is compressed ) Work done by spring on the mass W  Fx  x i x f  dx  kx x i x f  dx  x i x f  1 2 kx 2  1 2 f 2  1 2 i 2  by Newtons3rd law work done by mass on spring =  1 2 kx f 2  1 2 i 2    

9. Kinetic Energy and the Work-Energy theorem Consider constant net force F acting over distance d W net  Fd  mad using the kinetic equations of motion and noting that a  v f  v i t, we obtain  m v f  v i t     1 2 v f  v i  t      1 2 m v f 2  1 2 m v i 2 i.e. W net  1 2 m v f 2  1 2 m v i 2 Define Kinetic Energy K  1 2 m v 2  W net  K f  K i  K

10. Work - Energy theorem is valid for variable force W net  F(x)dx x i x f   ma(x)dx  m d v dt dx x i x f  x i x f   m d v dt dx x i x f   m d v v x i x f   m v d v v i v f   1 2 m v f 2  1 2 m v i 2  K

11. Power Definition P  dW dt For a force F that is constant over a displacement d P  dF  d  dt  dF  d  F  dd  P  F  v SI units of power P   F  v   W  T   J s  Watts(W )