General Physics 101 PHYS Dr. Zyad Ahmed Tawfik Email : zmohammed@inaya.edu.sa Website : zyadinaya.wordpress.com

Lecture No. 6 Work and Energy

Lecture No.6 Energy and work Vectors

Energy Symbol: E Units: J, Joule

Energy Energy Approach Energy definition Is the ability to do work. The energy approach use to describing motion of valid system, such as a car or a human body

Energy Every physical process that occurs in the Universe and every chemical reaction inside the human body involves energy, and this energy transfer or transformation from state to another state.

Types of Energy

Electrical Energy

Heat Energy

Chemical Energy Chemical Energy is energy stored in the bonds of chemical compounds (atoms and molecules). This type of energy available in nature, and the most important types of oil, coal, natural gas and wood

Electromagnetic Energy

2- Work

Work Work is the energy transferred to a system by a force that acts on it. Units of Work The unit of work is a joule (J) 1 joule = 1 Newton . 1 meter J = N · m

J = N · m Units of Work The unit of work is a joule (J) 1 joule = 1 Newton . 1 meter J = N · m

Work Work = The Scalar Dot Product between Force F and Displacement d. So that means if you apply a force on an object and it covers a displacement you have supplied ENERGY or done WORK on that object. W = F . d

Calculate work done on an object

1- With out angle In this case it means that F and d MUST be parallel. To ensure that they are parallel we add the cosine on the end. Apply FORCE Displacement So the equation used to calculate the work(W) in this case it: W= F . d

Also with friction force The equation used to calculate the work(W) in this case it: W= Ff . d This signal(-) Because in this case the friction force in the opposite direction for displacement Friction FORCE Displacement

Example 1,with out angle W = 50 · 10 = 500 J A 50 N horizontal force is applied to an object over a distance of 10 m. Find the amount of work done. Solution: since, W = F . d W = 50 · 10 = 500 J Object 50 N 10 m

With angle In the figure above, we see the woman applying a force at an angle theta q. Only the Horizontal component actually causes the box to move. and thus imparts energy to the box

W = F . d cos q So the force analysis into two components Fx and Fy Where Fx = F cos q and Fy = F sin q This vertical component (F sin q) is not doing any work on the box ( W=0) because it is perpendicular to the displacement). And also it is not parallel to the displacement. So in this case, the work done given by W = F . d cos q

Example1, with angle F sin F F cos A 50 N sloping force is applied to an object with angle 50 degree over a distance of 10 m. Find the amount of work done. Solution: since, W = F . d cos  W = 50 · 10 cos 50 = 500 x 0.642 = 321.39 J F sin F  F cos Object x

W= 50 . 5 cos30 W= 216.5 J Example2, with angle Suppose the woman in the figure above applies a 50 N force to a box at an angle of 30 degrees above the horizontal. She manages to pull the box 5 meters. Calculate the WORK done by the woman on the box? Solution: since, W = F . d cos  W = F . d cos  W= 50 . 5 cos30 W= 216.5 J

3- Kinetic Energy and Work Vectors

K = ½ mv2 (1) What is Kinetic Energy? Kinetic Energy is the energy of a particle due to its motion K = ½ mv2 (1) K is the kinetic energy m is the mass of the particle v is the speed of the particle

Work and Kinetic Energy Consider an object of mass m subjected to a constant force F , the object moves a distance x parallel to F. show figure Let, v0 is the initial velocity and Vf is the final velocity.

Work and Kinetic Energy Since, vf2 = v02 + 2 a d Multiplying by m/2, this becomes ½ mvf2 = ½ m v02 + m a d But F = m a, Also W = F d = m a d Then, ½ mvf2 = ½ m v02 + W But the low of Kinetic Energy(K= ½ mv2 ) so , Kf = K0 +W (1)

Work and Kinetic Energy Kf = K0 + W _______(1) Where Kf is the final kinetic energy K0 is the initial kinetic energy The equation (1) say (The final kinetic energy of an object = initial kinetic energy + total work done. Note-1, work and kinetic energy have the same dimensions and units

Work and Kinetic Energy Also from equation (1) we can Wright W = Kf - K0 _______(2) The equation (3) say (total work done= The final kinetic energy of an object - initial kinetic energy) _____________________________________________ But the equation 2 can be written in this format W =½ mvf2 - ½ m v02 -------------------- (3) where k= ½ m v2 But DK = ½ mvf2 - ½ mvo2 so w= DK ---------(5) The equation (5) say “the net WORK done on an object is equal to the change in kinetic energy of the object."

Example1, Suppose the woman in the figure above applies a 50 N force to 25-kg a box at an angle of 30 degrees above the horizontal. She manages to pull the box 5 meters. Calculate the WORK done by the woman on the box The speed of the box after 5 meters if the box started from rest. Solution: a) W = F . d cos  so W= 50 . 5 cos30 W= 216.5 J ____________________________________________________ b) w= DK = ½ mv2 where m=25 and w=216.5 216.5 = 1/2 . 25 v2 v= 4.16 m/s

4- Potential Energy Vectors

Potential Energy potential energy is energy stored in an object. Gravity gives potential energy to an object.   This potential energy is a result of gravity pulling downwards. Potential Energy means the work done by gravity on the object. The potential energy of the object is depend of the mass of the object (m), gravitational acceleration of the earth (g=9.8 m/sec2) and objective height above earth's surface (h). The formula for potential energy (U) due to gravity is U = m.g.h P.E. = mass x height x gravity

Example 1 What is the P.E. of a 50-kg person at a height of 480 m? Solution U = m.g.h M=50kg , g=10 and h= 480 U = m.g.h = (50 kg)(10 m/s2)(480 m)=240000 J U=240 K J

Example 2 Find the potential energy of 20 Kg mass child sitting on a roof 10 m above the ground.? Solution U = m.g.h M=20kg , g=10 and h= 10 U = m.g.h = (20 kg)(10 m/s2)(10 m)=1960 J U=2000 K J

Potential Energy and Kinetic energy Total mechanical energy = kinetic energy + potential energy E = K + u Potential Energy and Kinetic energy

5-law of conservation of mechanical energy Vectors

The law of conservation of mechanical energy The law of conservation of mechanical energy states: Energy cannot be created or destroyed, only transformed! Energy Before Energy After Am I moving? If yes, Ko Am I above the ground? If yes, Uo Am I moving? If yes, K Am I above the ground? If yes, U

Conservation of Energy In Figure A, a pendulum is released from rest at some height above the ground position. It has only potential energy. B In Figure B, a pendulum is still above the ground position, yet it is also moving. It has BOTH potential energy and kinetic energy. C In Figure C, a pendulum is at the ground position and moving with a maximum velocity. It has only kinetic energy. D In Figure D, the pendulum has reached the same height above the ground position as A. It has only potential energy.

Energy consistently changes forms

Energy consistently changes forms Am I above the ground? Am I moving? NO, h = 0, U = 0 J Yes, v = 8 m/s, m = 60 kg Position m v U K ME 1 60 kg 8 m/s (= U+K) 0 J 1920 J 1920 J

Energy consistently changes forms Energy Before = Energy After KO = U + K = (60)(9.8)(1) + (.5)(60)v2 1920= 588 + 30v2 = 30v2 44.4 = v2 v = 6.66 m/s Position m v U K ME 1 60 kg 8 m/s 0 J 1920 J 2 6.66 m/s 588 J 1332 J 1920 J

Energy consistently changes forms Am I moving at the top? No, v = 0 m/s EB = EA Using position 1 Ko = U = mgh 1920 =(60)(9.8)h h = 3.27 m Position m v U K ME 1 60 kg 8 m/s 0 J 1920 J 2 6.66 m/s 588 J 1332 J 3 0 m/s 1920 J 0 J

6- Power The Laws of Motion

Power Power is the time rate of energy transfer. The Power is given by It is the amount of energy consumed per unit time Power is the time rate of energy transfer. The Power is given by W F . d P = = t t Units of Power Where the unit of work(W) is joule and unit of time(t) is second. So The unit of power is a Watt where 1 watt = 1 joule / second

P = 33.33 watt Example 1 Solution: since, = = = W F . d 100 x 20 P A 100 N force is applied to an object in order to lift it a distance of 20 m within 60 s. Find the power. Solution: since, = = = P = 33.33 watt W F . d 100 x 20 P t t 60

Power Consumed: P = 2240 W Example 2 Solution: since, = = = = What power is consumed in lifting a 70-kg robber 1.6 m in 0.50 s? Solution: since, = = = = h =d and F = mg W F . h m.g.h 70 x10x 1.6 P t t t .50 Power Consumed: P = 2240 W

Thank You for your Attention