11/1/20151 Norah Ali Al Moneef king Saud university

2 W = F Δx cos  Work = the product of force and displacement, times the cosine of the angle between the force and the direction of the displacement. How much work is done pulling with a 15 N force applied at 20 o over a distance of 12 m? W = FΔx cos  W = 15 N (12 m) (cos 20) = 169J 20 o 15 N W = F. Δx units: N.m or joule, j 11/1/2015 Norah Ali Al Moneef king Saud university

+ve work: Force acts in the same direction as the displacement.  object gains energy  ve work: Force acts in the opposite direction as the displacement.  object loses energy Sign of work:  cos 90  = 0  No work is done on the load When force F is at right angles to displacement s (F ⊥ s): Δx  F Δx cos  = 0 No work is done if: F F = 0 or Δx Δx = 0 or  = 90 o 11/1/20153 Norah Ali Al Moneef king Saud university

A 1500 kg car moves down the freeway at 30 m/s K.E. = ½  (1500 kg)  (30 m/s) 2 = 675,000 kg·m 2 /s 2 = 675 kJ A 2 kg fish jumps out of the water with a speed of 1 m/s K.E. = ½  (2 kg)  (1 m/s) 2 = 1 kg·m 2 /s 2 = 1 J Calculate the work done by a child of weight 300N who climbs up a set of stairs consisting of 12 steps each of height 20cm. work = force x distance the child must exert an upward force equal to its weight the distance moved upwards equals (12 x 20cm) = 2.4m work = 300 N x 2.4 m= 720 J example 11/1/20154 Norah Ali Al Moneef king Saud university

6-2 Kinetic Energy Same units as work Remember the Eq. of motion Multiply both sides by m, When work is done on a system and the only change in the system is in its speed, the work done by the net force equals the change in kinetic energy of the system. The speed of the system increases if the work done on it is positive The speed of the system decreases if the net work is negative W = F  x W =  K Work-Energy Theorem 11/1/20155 Norah Ali Al Moneef king Saud university

6 Kinetic Energy – the energy of motion KE = ½ m v 2 units: kg (m/s) 2 = (kgm/s 2 ) m = Nm = joule Work Energy Theorem: W =  KE Work = the change in kinetic energy of a system. Note: v = speed NOT velocity. The direction of motion has no relevance to kinetic energy. 11/1/2015 Norah Ali Al Moneef king Saud university

Factors Affecting Kinetic Energy The kinetic energy of an object depends on both its mass and its velocity. Kinetic energy increases as mass increases. For example, think about rolling a bowling ball and a golf ball down a bowling lane at the same velocity, as shown in Figure 2. The bowling ball has more mass than the golf ball. Therefore, you use more energy to roll the bowling ball than to roll the golf ball. The bowling ball is more likely to knock down the pins because it has more kinetic energy than the golf ball. 11/1/20157 Norah Ali Al Moneef king Saud university

Kinetic energy also increases when velocity increases. For example, suppose you have two identical bowling balls and you roll one ball so it moves at a greater velocity than the other. You must throw the faster ball harder to give it the greater velocity. In other words, you transfer more energy to it. Therefore, the faster ball has more kinetic energy. Figure 2Kinetic Energy Kinetic energy increases as mass and velocity increase. Predicting In each example, which object will transfer more energy to the pins? Why? 11/1/20158 Norah Ali Al Moneef king Saud university

Calculating Kinetic Energy There is a mathematical relationship between kinetic energy, mass, and velocity. Example : A 6 kg bowling ball moves at 4 m/s. a) How much kinetic energy does the bowling ball have? b) How fast must a 2.5 kg tennis ball move in order to have the same kinetic energy as the bowling ball? K = ½ m b v b ² K = ½ (6 kg) (4 m/s)² KE = 48 J KE = ½ m t v t ² v = √2 K/m = 6.20 m/s 11/1/20159 Norah Ali Al Moneef king Saud university

Do changes in velocity and mass have the same effect on kinetic energy? No—changing the velocity of an object will have a greater effect on its kinetic energy than changing its mass. This is because velocity is squared in the kinetic energy equation. For instance, doubling the mass of an object will double its kinetic energy. But doubling its velocity will quadruple its kinetic energy. 11/1/ Norah Ali Al Moneef king Saud university

Calculate the speed of a car of mass 1200kg if its kinetic energy is J K = ½ m v J = ½ x 1200kg x v = 600 x v ÷ 600 = v 2 25 = v 2 v =  25 speed = 5.0 ms -1 example 11/1/ Norah Ali Al Moneef king Saud university

example Calculate the braking distance a car of mass 900 kg travelling at an initial speed of 20 ms -1 if its brakes exert a constant force of 3 kN. k.E. of car = ½ m v 2 = ½ x 900kg x (20ms -1 ) 2 = ½ x 900 x 400 = 450 x 400 k = J The work done by the brakes will be equal to this kinetic energy. W = F Δx J = 3 kN x Δx = 3000 x Δx Δx = / 3000 braking distance = 60 m 11/1/ Norah Ali Al Moneef king Saud university

An object does not have to be moving to have energy. Some objects have stored energy as a result of their positions or shapes. When you lift a book up to your desk from the floor or compress a spring to wind a toy, you transfer energy to it. The energy you transfer is stored, or held in readiness. It might be used later when the book falls to the floor Stored energy that results from the position or shape of an object is called potential energy. This type of energy has the potential to do work. 11/1/ Norah Ali Al Moneef king Saud university

Gravitational Potential Energy Potential energy related to an object’s height is called gravitational potential energy. The gravitational potential energy of an object is equal to the work done to lift it. Remember that Work = Force × Distance. The force you use to lift the object is equal to its weight. The distance you move the object is its height. You can calculate an object’s gravitational potential energy using this formula: Gravitational Potential Energy = Weight x Height change in Gravitational potential energy. = mass x gravitational field strength x change in height ΔU = m g Δh 11/1/ Norah Ali Al Moneef king Saud university

Gravitational potential energy: -Potential energy only depends on y (height) and not on x (lateral distance) -MUST pick a point where potential energy is considered zero! 11/1/ Norah Ali Al Moneef king Saud university

Calculate the change in Gravitational potential energy (g.p.e). when a mass of 200 g is lifted upwards by 30 cm. (g = 9.8 Nkg -1 ) ΔU = m g Δh = 200 g x 9.8 Nkg -1 x 30 cm = kg x 9.8 Nkg -1 x 0.30 m change in g.p.e. = 0.59 J example 11/1/ Norah Ali Al Moneef king Saud university

Q: Calculate the gravitational potential energy in the following systems a. a car with a mass of 1200 kg at the top of a 42 m high hill (1200 kg)( 9.8m/s/s)(42 m) = 4.9 x 10 5 b. a 65 kg climber on top of Mt. Everest (8800 m high) (65 kg) (9.8m/s/s) (8800 m) = 5.6 x 10 6 J c. a 0.52 kg bird flying at an altitude of 550 m (.52 kg) (9.8m/s/s)(550) = 2.8 x 10 3 J example 11/1/ Norah Ali Al Moneef king Saud university

Conservation of Energy Law of conservation of energy - energy cannot be created or destroyed closed system: - all energy remains in the system - nothing can enter or leave open system: - energy present at the beginning of the system may not be at present at the end 11/1/ Norah Ali Al Moneef king Saud university

Conservation of Energy When we say that something is conserved it means that it remains constant, it doesn’t mean that the quantity can not change form during that time, but it will always have the same amount. Conservation of Mechanical Energy: ME i = ME f initial mechanical energy = final mechanical energy If the only force acting on an object is the force of gravity: K o + U o = K + U ½ mv o ² + mgh o = ½ mv² + mgh 11/1/ Norah Ali Al Moneef king Saud university

At a construction site a 1.50 kg brick is dropped from rest and hit the ground at a speed of 26.0 m/s. Assuming air resistance can be ignored, calculate the gravitational potential energy of the brick before it was dropped? K+U = K o +U o ½ mv 2 +mgh=½ mv 2 o + mgh o ½ mv 2 +0 = =o+ mgh o U 0 = mgh o = ½ (1.50 kg) (26.0m/s) 2 = 507 J example 11/1/ Norah Ali Al Moneef king Saud university

example A child of mass 40 kg climbs up a wall of height 2.0 m and then steps off. Assuming no significant air resistance calculate the maximum: (a) gravitational potential energy (gpe) of the child (b) speed of the child g = 9.8 Nkg -1 (a) maximum gravitational potential energy ( max gpe) occurs when the child is on the wall Δ U = mgΔh = 40 x 9.8 x 2.0 max gpe = 784 J (b) max speed occurs when the child reaches the ground ½ m v 2 = m g Δh ½ m v 2 = 784 J v 2 = (2 x 784) / 40 v 2 = 39.2 v =  39.2 max speed = 6.3 ms -1 11/1/ Norah Ali Al Moneef king Saud university

A 10 kg shopping cart is pushed from rest by a 250 N force against a 50 N friction force over 10 m distance. m = 10.0 kg v i = 0 F p = N F k = 50.0 N Δx = 10.0 m FpFp FkFk FNFN FgFg A. How much work is done by each force on the cart? W g = 0 W N = 0 W p = F p Δx cos  = (250.0 N)(10.0 m)cos 0 W p = 2500 J = -500 J W k = F k Δx cos  = (50.0 N)(10.0 m)cos 180 = -500 J example 11/1/ Norah Ali Al Moneef king Saud university

B. How much kinetic energy has the cart gained? W net = ∆KE W p + W k = KE f - KE i 2500 J J = KE f - 0 KE f = 2000J C. What is the carts final speed? KE = 1/2 m v 2 v = √((2KE)/(m)) v = √((2(2000 J))/(10.0 kg)) v = 20 m/s 11/1/ Norah Ali Al Moneef king Saud university

Nonconservative Forces A nonconservative force does not satisfy the conditions of conservative forces Nonconservative forces acting in a system cause a change in the mechanical energy of the system The work done against friction is greater along the brown path than along the blue path Because the work done depends on the path, friction is a nonconservative force 11/1/ Norah Ali Al Moneef king Saud university

On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial speed of 2.2 m/s. How far does it travel if the coefficient of kinetic friction between the sled and the ice is.10? m = 10.0 kg v i = 2.2 m/s v f = 0 m/s µ k =.10 W = ∆KE FΔx= K f - K i µ k = F k / F N F k = µ k (mg) µ k (-mg) Δx = 1/2 m v f 2 - 1/2 m v i 2 µ k (-mg) Δx = - 1/2 m v i 2 Δx = (- 1/2 m v i 2 ) / µ k (-mg) Δx = (- 1/2 (10.0 kg) (2.2 m/s) 2 ) / (.10 (-10.0 kg) (9.8 m/s 2 )) Δx = 2.47 m example 11/1/ Norah Ali Al Moneef king Saud university

A 20-g projectile strikes a mud bank, penetrating a distance of 6 cm before stopping. Find the stopping force F if the entrance velocity is 80 m/s. W =Δ K F Δ x cosθ = K – K 0 F Δ x cosθ = ½ mv² - ½ mv o ² F (0.06 m) cos = (0.02 kg)(80 m/s) 2 F (0.06 m)(-1) = -64 J F = 1067 N Work to stop bullet = change in K.E. for bullet Example W = ½ mv² - ½ mv o ² = 0 11/1/ Norah Ali Al Moneef king Saud university

A bus slams on brakes to avoid an accident. The tread marks of the tires are 80 m long. If μk = 0.7, what was the speed before applying brakes? Work = F Δx (cos θ) f = μ k.N = μ k mg Work = - μ k mg Δx ΔK = 1/2 mv 2 - ½ mv o 2 W = ΔK v o = (2μ k g Δ x) ½ example 11/1/ Norah Ali Al Moneef king Saud university

Power is the rate at which work is done. Power is the rate of energy transfer by any method. The SI unit of power is the watt, W 1 watt = 1 Joule/s 1000 watts = 1 kW Work = force · distance Power = force·distance/time Power = force·velocity P = F·v 11/1/ Norah Ali Al Moneef king Saud university

Calculate the power of a car engine that exerts a force of 40 kN over a distance of 20 m for 10 seconds. W = FΔx = 40 kN x 20 m = x 20 m = J example P = ΔW / Δt = J / 10 s power = W In 1935, R. Goddard, the America rocket pioneer, launched several A-series sounding rockets. Given that the engine had a constant thrust of 890 N, how much power did it transfer to the rocket while traveling at its maximum speed of m/s. P = Fv = 890N x m/s = 279 kW. 11/1/ Norah Ali Al Moneef king Saud university

30 example Together, two students exert a force of 825 N in pushing a car a distance of 35 m. A. How much work do they do on the car? W = F Δx = 825N (35m) = B. If the force was doubled, how much work would they do pushing the car the same distance? W = 2F Δx = 2(825N)(35m) = 11/1/2015 Norah Ali Al Moneef king Saud university

31 A rock climber wears a 7.5 kg backpack while scaling a cliff. After 30.0 min, the climber is 8.2 m above the starting point. a.How much work does the climber do on the backpack? W = Fd = m g Δx= 7.5kg(9.8m/s 2 )(8.2m) b.How much power does the climber expend in this effort? P = W = 603 J = 0.34 watt t 1800 s example 11/1/2015 Norah Ali Al Moneef king Saud university

Example : A 200 kg curtain needs to be raised 8m. in as close to 5 s as possible. You need to decide among three motors to buy for this, each motor cost a little more the bigger the power rating. The power rating for the three motors are listed as 1.0 kw, 3.5 kw and 5.5 kw. Which motor is the best for the job? Given: m = 200 kg Δx = 8 m ∆t = 5s Unknown: Power and work W = F·Δx W = m·g·h W = (200 kg)·(9.81 m/s²)·(8 m) = 15,696 Joules P = W/∆t P = 15,696 J / 5 s= 3,139 watts 11/1/ Norah Ali Al Moneef king Saud university

example Calculate the power of a car that maintains a constant speed of 30 ms -1 against air resistance forces of 2 kN As the car is travelling at a constant speed the car’s engine must be exerting a force equal to the opposing air resistance forces. P = F v = 2 kN x 30 ms -1 = N x 30 ms -1 power = 60 kW 11/1/ Norah Ali Al Moneef king Saud university

What power is consumed in lifting a 70-kg robber 1.6 m in 0.50 s? A 100-kg cheetah moves from rest to 30 m/s in 4 s. What is the power? the work is equal to the change in kinetic energy: 11/1/ Norah Ali Al Moneef king Saud university

Summary If the force is constant and parallel to the displacement, work is force times distance If the force is not parallel to the displacement, The total work is the work done by the net force: SI unit of work: the joule, J Total work is equal to the change in kinetic energy: where Power is the rate at which work is done: SI unit of power: the watt, W 11/1/ Norah Ali Al Moneef king Saud university