Presentation is loading. Please wait.

Presentation is loading. Please wait.

Work.  The product of the magnitudes of the component of a force along the direction of displacement and the displacement.  Units-Force x Length  N.

Similar presentations


Presentation on theme: "Work.  The product of the magnitudes of the component of a force along the direction of displacement and the displacement.  Units-Force x Length  N."— Presentation transcript:

1 Work

2  The product of the magnitudes of the component of a force along the direction of displacement and the displacement.  Units-Force x Length  N x m = Nm = J (Joules)  W=Fd (force times displacement)

3  If an object is not moved, NO WORK is done.  If you hold a heavy object, your body has work being done inside, but no work is done on the object.  For W = Fd, the force must be the component in the same direction as the displacement.

4  You push on a van directly horizontal  W = (your force) x (distance moved horizontally)  You push down on a bumber at an angle  Find the force of the horizontal component  Use cos θ ; so W = Fd cosθ

5  You pull a crate by attaching a rope at an angle of 25 ⁰. You then pull with 100 N, and it moves 10m. Calculate the work done?  Given:  F=100 N  Θ=25 ⁰  d=10 m  Work:  W = Fd cosθ  W = (100 N)(10 m) cos(25 ⁰ )  W=900 J  Homework: Practice 5A Page 162 #1-4

6  Problem #1  Given:  F= 5.00 x 10 3 N  d= 3.00Km=3000 m  Work:  W=Fd  W=(5.00 x 10 3 N)(3000m)  W=1.50 x 10 7 J

7  Problem #2  Given:  F= 350 N  d= 2.00m  Work:  W=Fd  W=(350 N)(2.00 m)  W=7.0 x 10 2 J

8  Problem #3  Given:  F= 35 N  d= 50.0 m  Θ= 25 ⁰  Work:  W=Fd cosθ  W=(35 N)(50.0 m)cos(25 ⁰)  W=1.6 x 10 3 J (1600 J)

9  Not what you might think  Scalar, but can be + or –  Positive-component force in same direction as displacement.  Negative- If opposite (Kinetic Friction Force)  Cos θ is negative for angles greater than 90 ⁰ but less than 270⁰  When speed is changed by work:  If sign is positive, increases  If sign is negative, decreases  Positive Work - work done ON the object  Negative Work -work done BY the object

10  Energy associated with an objects motion  Does not depend on direction; scalar  Depends on mass and speed  KE = (1/2) mv 2  m=mass  v=velocity  Units  Kg (m/s) 2  Kg m 2 /s 2 = Joule (J)

11  Stored energy; depends on properties of the object AND it position (reference to environment)  Two Types  Gravitational (PEg)  Elastic (PEelastic)  Total PE= PEg + PE (elastic)

12  The energy associated with an object due to the object’s position relative to a gravitational source.  PEg = mgh  m=mass  g=9.81m/s 2  h=height (In measuring height, choose arbitrary zero level.)  Units=Joules (J)  Note: Gravity and Free-fall acceleration are not properties of the object

13  Potential energy in a stretched or compressed elastic object.  PE (elastic) = (1/2) Kx 2  k-spring constant, specific to the elastic object unit N/m  x-distance stretched/compressed from resting  Unit-N/m

14  A 700kg stuntman is attached to a bungee cord with an unstretched length of 15.0m. He jumps off a bridge from a height of 50.0m. When he finally stops, the cord has a stretched length of 44.0m. Disregard weight of cord K = 71.8 N/m. What is total PE relative to the water when the man stops falling?

15  Given:  m=70.0 Kg  h=50.0 m – 44.0 m = 6.0m  x=44.00 m - 15.0 m = 29.0 m  PE = 0 at River Level  Work:  PEg = mgh = (70.0)(9.81)(6.0) = 4100 J  PE(elastic) = (1/2)kx 2 = (1/2)(71.8)(29.0) 2 = 30,200J  PE total = PEg + PE (elastic)  PE = 34, 300 J

16  Mechanical Energy-the sum of kinetic energy and all forms of potential energy  ME = KE + PE  In the absence of friction, ME remains constant; CONSERVATION of MECHANICAL ENERGY  ME i = ME f (no friction)  1. If only force is gravity:  1/2mv i 2 + mgh i = 1/2mv f 2 + mgh f  2. What if you had gravity and a spring?

17  PE = mgh  hi = 1.1 m  hf = 0 m  m= 2.0Kg  g = 9.81 m/s 2  KE = ½ mv 2  Vf= ?  Vi = 0

18  PEgi + KEi = PEgf +KEf  mghi + (1/2)mv 2 = mghf + (1/2)mvf 2  (2.0)(9.81)(1.1) + 0 = 0 + (1/2)(2.0)(Vf) 2  Vf = √(21.56)  Vf = 4.7 m/s


Download ppt "Work.  The product of the magnitudes of the component of a force along the direction of displacement and the displacement.  Units-Force x Length  N."

Similar presentations


Ads by Google