1 Momentum, p is mass times velocity: p = m v vector! unit: (p) = kg m/s Newton s second law: Force = time rate of change of momentum Net force F will produce change in momentum Δp of the object on which it acts during time interval Δt. If mass remains constant (well known form of Newton s second law) Impulse of the force F: F t (action of a force F over time Δt ) units: (F t) = Ns Ns = kg m/s Impulse F t acting on an object will produce change in momentum Δp of that object: velocity, force and momentum are vectors, therefore direction matters!!!!!!! F t = p = mv mu You probably have an experience or you can imagine or we can try it: If you hold a racket in your hand and a ball is coming toward you, you can choose ONLY to stop a ball or to BOUNCE it back. It is harder to bounce it back and your arm will hurt more. So it seems that you have to exert more force in case that you push it back. How can we explain that using physics? Let s say that we have a camera and motion detector. Camera can give us the time of contact of the ball and racket (Δt) and the motion detector measures the speed of the ball before and after contact. We want to find the force exerted by the racket (you). Let s say: the ball of 0. kg has speed before contact 5 m/s and the direction is to the right. The camera says that the time of contact (in physics we say the time of interaction) is 0.5 s. You managed to tilt the racket in such a way that you stop the ball and the ball drops down. How much force did you exert on the ball? How much force did the ball exert on the racket (or how much it hurts you?) PHYSICS: Impulse, FΔt, exerted on the ball will change momentum of the ball FΔt = Δp = mv mu force x time interval = final momentum (after contact) initial momentum (before contact) momentum p = mv is a vector because v in this formula is a vector (speed + direction) direction matters, so we have to have a coordinate system coordinate system: u = 5 m/s to the right v = 0. Let right be positive, then u = +5 m/s FΔt = Δp = mv mu F(0.5) = (0.)(0) (0.)(5) 0.5F = 1 F = N or F = N to the left you have to exert the force of N to the left to stop the ball; The ball exerted N to the right on you. And you felt it. Now, let s imagine that you not only stopped the ball but you pushed back it the left with speed 4 m/s. How much force did you exert on the ball? coordinate system: u = 5 m/s to the right v = 4 m/s to the left. Let right be positive, then u = +5 m/s and v = 4 m/s FΔt = Δp = mv mu F(0.5) = (0.)(-4) (0.)(5) 0.5F = 1.8 F = 3.6N or F = 3.6 N to the left you have to exert the force of 3.6N to the left to bounce the ball The ball exerted 3.6N to the right on you. It hurts even more. In short, in pictures: 1. case: How much force did you exert on the ball? FΔt = Δp = mv mu F(0.5) = (0.)(0) (0.)(5) 0.5F = 1 F = N or F = N to the left. case: How much force did you exert on the ball? FΔt = Δp = mv mu F(0.5) = (0.)(-4) (0.)(5) 0.5F = 1.8 F = 3.6N or F = 3.6 N to the left you have to exert the force of 3.6N to the left to bounce the ball. The ball exerted 3.6N to the right on you.
1. Which one of the following is true concerning momentum? a. Momentum is a force. b. Momentum is a scalar quantity. c. The SI unit of momentum is kg m /s. d. The momentum of an object is always positive. e. Momentum and impulse are measured in the same units. e. Momentum and impulse are measured in the same units Ns = (kg m/s )s = kg m/s. A stunt person jumps from the roof of a tall building, but no injury occurs because the person lands on a large, air-filled bag. Which one of the following best describes why no injury occurs? a. The bag provides the necessary force to stop the person. b. The bag reduces the impulse to the person. c. The bag increases the amount of time the force acts on the person and reduces the change in momentum. d. The bag decreases the amount of time during which the momentum is changing and reduces the average force on the person. e. The bag increases the amount of time during which the momentum is changing and reduces the average force on the person. e. The bag increases the amount of time during which the momentum is changing and reduces the average force on the person. 3. Jennifer is walking at 1.63 m/s. If Jennifer weighs 583 N, what is the magnitude of her momentum? W = mg m = W/g = 583 /9.80 = 59.5 kg p = mv = (59.5 kg)( 1.63 m/s) = 97.0 kg m/s 4. What impulse will give a.0-kg object a momentum change of + 50 kg m/s? a.+5 N s b. 5 N s c. +50 N s d. 50 N s e. +100 N s c. change of momentum = impulse given to that object: p = F t F t = + 50 kg m/s = +50 N s 5. A 1.0-kg ball has a velocity of 1 m/s downward just before it strikes the ground and bounces up with a velocity of 1 m/s upward. What is the change in momentum of the ball? choose positive direction. I chose upward to be positive. - 1 kg m/s Δp m(v u) 1.0 1-4 positive direction means upward, so Δp 4kg m/s, upward 6. Two bodies of unequal mass, placed at rest on a frictionless surface, are acted on by equal horizontal forces for equal times. Just after these forces are removed, the body of greater mass will have: a. the greater speed b. the greater acceleration c. the smaller momentum d. the greater momentum e. the same momentum as the other body e. Impulses (F t) acting on two bodies are equal, therefore changes in their momenta are equal. 7. A 0. kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds up at 0 m/s. The magnitude of the impulse due to the collision with the sidewalk is: impulse (F t) = momentum change ( p). choose positive direction. I chose upward to be positive. Δp pf - pi 4 kg m/s - (-6 kg m/s) 10kg m/s positive direction means upward, so impulse = 10 Ns, up 8. A 10-kg block of ice is at rest on a frictionless horizontal surface. A 1.0-N force is applied in an easterly direction for 1.0 s. During this time interval, the block: a. acquires a speed of 1 m/s b. moves 10 cm c. acquires a momentum of 1.0 kg m/s c. momentum change ( p) = impulse (F t) F t = 1.0 Ns so p = 1.0 kg m/s
3 9. A 0.065-kg tennis ball moving to the right with a speed of 15 m/s is struck by a tennis racket, causing it to move to the left with a speed of 15 m/s. If the ball remains in contact with the racquet for 0.00 s, what is the magnitude of the average force experienced by the ball? m = 0.065 kg v i = 15 m/s, right v f = 15 m/s, left t = 0.00s F =? F t = p F = p/ t left (final direction) is positive Δp 0. 065 (v u) 0. 065 15 - - 15 1. 95 kg m/s F = p/ t = 1.95 Ns/0.00 s F = 98 N 10. A bat strikes a 0.050-kg baseball so that its velocity changes by +30 m/s in 0.10 s. With what average force was the ball struck? m = 0.050 kg v = +30 m/s t = 0.10 s F =? F t = p F = p/ t F t = m v = 0.050 kg (+30 m/s) = 1.5 kg m/s F = 1.5 Ns/0.10 s F = 15 N 11. What velocity must a 1340 kg car have in order to have the same momentum as a 680 kg truck traveling at a velocity of 15 m/s to the west? P 1 = p m 1 v 1 = m v (1340 kg) v 1 = (680 kg)( 15 m/s, west) v 1 = 30 m/s, west 1. A force is applied to stop a moving shopping cart. Increasing the time interval over which the force is applied a. requires a greater force. b. has no effect on the force needed. c. requires a smaller force. d. requires the same force. c. requires a smaller force change of momentum = impulse given to that object (F t), therefore the longer time interval has to be multiplied with the smaller force in order to get the same change in momentum 13. A 6.0 x 10 kg tennis ball moves at a velocity of 1 m/s. The ball is struck by a racket, causing it to rebound in the opposite direction at a speed of 18 m/s. What is the change in the ball s momentum? p =? left (final direction) is positive Δp m(v u) - 6.0 x 10 kg 18m/s - - 1m/s 1. 8 kg m/s Δp 1.8 kg m/s, left or Δp - 1.8 kg m/s, right 14. A rubber ball with a mass of 0.30 kg is dropped onto a steel plate. The ball s velocity just before impact is 4.5 m/s and just after impact is 4. m/s. What is the change in the ball s momentum? p =? p i = 1.35 kg m/s, downward p f = 1.6 kg m/s, upward p = p f p i = 1.6 (-1.35) =.6 kg m/s p =.6 kg m/s, upward = -.6 kg m/s, downward 15. A 0. baseball if pitched with a velocity of 40 m/s and is then batted to the pitcher with a velocity of 60 m/s. What is the magnitude of change in the ball s momentum? p =? left (final direction) is positive 0.kg 40m/s - - 60m/s kg m/s Δp m(v u) 0 Δp 0 kg m/s, left or Δp - 0kg m/s, right
4 Collisions and explosions If collision or explosion happens in an isolated system momentum is conserved. Law of Conservation of Momentum The total momentum of a system of interacting particles is conserved - remains constant, provided there is no resultant external force. Such a system is called an isolated system. p after collision = p before collision where p is momentum of the system: p = p1 + p DIRECTION MATTERS Inelastic collision: momentum is conserved but KE is not conserved. Perfectly inelastic collision: objects after collision stick together To find how much of KE is lost in the system, subtract KE of the system after collision from KE of the system before the collision. Elastic collision both momentum and kinetic energy are conserved.. 1. A swimmer with a mass of 75 kg dives off a raft with a mass of 500 kg. If the swimmer s speed is 4 m/s immediately after leaving the raft, what is the speed of the raft? mu = m 1 v 1 + m v 0 = 500 v 1 + 300 v 1 = 0.6 m/s negative sign means in the direction opposite to the swimmer. An astronaut with a mass of 70.0 kg is outside a space capsule when the tether line breaks. To return to the capsule, the astronaut throws a.0 kg wrench away from the capsule at a speed of 14 m/s. At what speed does the astronaut move toward the capsule? mu = m 1 v 1 + m v 0 = 70 v 1 + 8 v 1 = 0.4 m/s negative sign means in the direction opposite to the capsule 3. A bullet with a mass of 5.00 x 10 3 kg is loaded into a gun. The loaded gun has a mass of 0.5 kg. The bullet is fired, causing the empty gun to recoil at a speed of.1 m/s. What is the speed of the bullet?. mu = m 1 v 1 + m v 0 = 0.005 v 1 + 1.1 v 1 = 0 m/s negative sign means in the direction opposite to the gun 4. Two carts with masses of 1.5 kg and 0.7 kg, respectively, are held together by a compressed spring. When released, the 1.5 kg cart moves to the left with a velocity of 7 m/s. What is the velocity of the 0.7 kg cart? (Disregard the mass of the spring.) mu = m 1 v 1 + m v 0 = 10.5 + 0.7 v v = 15 m/s v = 15 m/s to the right
5 5. A bowling ball with a mass of 7.0 kg strikes a pin that has a mass of.0 kg. The pin flies forward with a velocity of 6.0 m/s, and the ball continues forward at 4.0 m/s. What was the original velocity of the ball? m 1 u 1 + m u = m 1 v 1 + m v 7.0 u 1 + 0 = 8 + 1 u 1 = 5.7 m/s u 1 = 5.7 m/s to the right 6. A clay ball with a mass of 0.35 kg strikes another 0.35 kg clay ball at rest, and the two balls stick together. The final velocity of the balls is.1 m/s north. What was the first ball s initial velocity? m 1 u 1 + m u = (m 1 + m ) v 0.35 u 1 + 0 = 1.47 u 1 = 4. m/s u 1 = 4. m/s to the north 7. An 80-kg astronaut carrying a 0-kg tool kit is initially drifting toward a stationary space shuttle at a speed of m/s. If she throws the tool kit toward the shuttle with a speed of 6 m/s as seen from the shuttle, her final speed is (m 1 + m ) u = m 1 v 1 + m v 00 = 80 v 1 + 10 v 1 = 1 m/s v 1 = 1 m/s, toward the shuttle 8. A 58.5-kg astronaut is floating toward the front of her stationary ship at 0.15 m/s, relative to the ship. She wishes to stop moving, relative to the ship. She decides to throw away the.50-kg book she's carrying. What should the speed and direction of the book be to achieve her goal? (m 1 + m ) u = m 1 v 1 + m v 9.15 = 0 +.50 v v 1 = 3.7 m/s v 1 = 3.7 m/s, toward the front of the ship 9. A 3.0-kg cart moving to the right with a speed of 1.0 m/s has a head-on collision with a 5.0-kg cart that is initially moving to the left with a speed of m/s. After the collision, the 3.0-kg cart is moving to the left with a speed of 1 m/s. What is the final velocity of the 5.0-kg cart? m 1 u 1 + m u = m 1 v 1 + m v 3.0-10.0 = -3.0 + 5 v v 1 = - 0.8 m/s v 1 = 0.8 m/s to the left 10. A 1000-kg car traveling east at 0 m/s collides with a 1500-kg car traveling west at 10 m/s. The cars stick together after the collision. What is their common velocity after the collision? m 1 u 1 + m u = (m 1 + m ) v v = m/s, east 0000 15000 = 500 v 3.0-10.0 = -3.0 + 5 v v = m/s
11. A 75-kg man is riding in a 30-kg cart at.0 m/s. He jumps off in such a way as to land on the ground with no horizontal velocity. The resulting change in speed of the cart is: 6. mu = m 1 v 1 + m v 10 = 0 + 30 v v = 7 m/s, right The resulting change in speed of the cart is 7 m/s m/s = 5.0 m/s 1. A 7.30-kg bowling ball strikes a 1.60-kg pin at rest head-on. Before the collision, the velocity of the ball is +6.00 m/s. After the collision, the velocity of the ball is +5.40 m/s. What is the velocity of the pin after the collision? m 1 u 1 + m u = m 1 v 1 + m v 43.8 + 0 = 39.4 + 1.6 v v = +.7 m/s 13. Batman (mass = 9.0 kg) jumps straight down from a bridge into a boat (mass = 4 kg) in which a criminal is fleeing. The velocity of the boat is initially +10.4 m/s. What is the velocity of the boat after Batman lands in it? Batman doesn t have horizontal velocity, only vertical; so for horizontal direction: Batman doesn t have horizontal velocity, only vertical; so for horizontal direction: m 1 u 1 + m u = (m 1 + m ) v (4)(10.4) = 514 v v = 0.1 m/s Work done by a constant force F exerted on an object through distance d is: W = Fd cos Ѳ (F d = F cos Ѳ) Work = (force)x(distance)(cos of the angle between force and distance) = (component of the force along direction of motion)x(distance) Work done by force F is positive when the force helps the motion (the force and direction of motion are generally in the same direction (cos θ = +) and negative when the force opposes the motion (the force and direction of motion are in generally opposite direction (cos θ = ). Work done by friction force is always negative (cos(180 0 ) = 1). W fr = F fr d. Work done by force F is zero if: the force is exerted but no motion is involved or there is no component of the force in the direction of motion ( the force is perpendicular to the direction of motion (cos 90 0 = 0) Gravitational Potential energy, PE = mgh Work done on an object by an applied force against gravitational force (when the net force is zero, so there is no acceleration) is stored as gravitational potential energy of the object. Elastic Potential energy, EPE = ½ kx Work done on a spring by applied force (F = kx) against the spring force (F = kx) (when the net force is zero, so there is no acceleration) is stored in stretched/compressed spring as elastic potential energy. A spring can be stretched or compressed. The same mathematics holds for stretched as for compressed springs. Kinetic energy KE = ½ mv Work Kinetic energy relationship: work done by net force changes kinetic energy W = KE = KE f KE i F fr d = ½ mv ½ mu When the net force and direction of motion are in generally the same directions, work is positive and KE is increasing. When the net force and direction of motion are in generally opposite directions, work is negative and KE is decreasing. ALL energies and work are measured in JOULES (J) 1J = 1Nm
7 1.A concrete block is pulled 7.0 m across a frictionless surface by means of a rope. The tension in the rope is 40 N; and the net work done on the block is 47 J. What angle does the rope make with the horizontal? W = Fd cos 47 = (40) (7.0) (cos ) = 8 0 3.Mike is cutting the grass using a human-powered lawn mower. He pushes the mower with a force of 45 N directed at an angle of 41 below the horizontal direction. Calculate the work that Mike does on the mower in pushing it 9.1 m across the yard. F = 45 N d = 9.1 m θ = 41 0 W = Fd cos θ = 310 J 3.A 5.00-kg block of ice is sliding across a frozen pond at.00 m/s. A 7.60-N force is applied in the direction of motion. After the ice block slides 15.0 m, the force is removed. The work done by the applied force is W = Fd cos = (7.60) (15.0) cos 0 0 = 114 J 4.A force of magnitude 5 N directed at an angle of 37 above the horizontal moves a 10-kg crate along a horizontal surface at constant velocity. How much work is done by this force in moving the crate a distance of 15 m? W = Fd cos = (5) (15) cos 37 0 = 300 J 5.A constant force of 5 N is applied as shown to a block which undergoes a displacement of 7.5 m to the right along a frictionless surface while the force acts. What is the work done by the force? W = Fd cos = (5) (7.5) cos 150 0 = -160 J The force is slowing down object 6. A 1.0-kg ball on the end of a string is whirled at a constant speed of.0 m/s in a horizontal circle of radius 1.5 m. What is the work done by the centripetal force (tension force) during one revolution? zero Joules tension force is pointing toward the center of the circle so it perpendicular to the direction of motion 7. Brenda carries an 8.0-kg suitcase as she walks 5 m along a horizontal walkway to her room at a constant speed of 1.5 m/s. How much work does Brenda do in carrying her suitcase? zero Joules Brenda s force is perpendicular to the direction of motion 8.The kinetic energy of a car is 8 x 10 6 J as it travels along a horizontal road. How much work is required to stop the car in 10 s? W = ΔKE = KE f KE i W = 8 x 10 6 J 9.How much energy is dissipated in braking a 1000-kg car to a stop from an initial speed of 0 m/s? dissipated energy energy that is converted into heat - is the work done by friction force. Friction is the only and the net force acting on the car, so W = ΔKE = KE f KE i = 0 ½ (100) (0) = 00 000 J ( negative means final KE is less then initial KE) 10. The kinetic energy of an 1100-kg truck is 4.6 x 10 5 J. What is the speed of the truck?. E = ½ m v v = E m v = 9 m/s 11. A 40-kg block is lifted vertically 0 meters from the surface of the earth. To one significant figure, what is the change in the gravitational potential energy of the block? PE = mgh = 8000 J
8 1. A 1500-kg elevator moves upward with constant speed through a vertical distance of 5 m. How much work was done by the tension in the cable? work is now stored as PE in the elevator PE = mgh = (1500)(10(5) = 375 000 J 13. A helicopter (m = 150 kg) is cruising at a speed of 5.0 m/s at an altitude of 185 m. What is the total mechanical energy (the sum of PE and KE) of the helicopter? ME = PE + KE = mgh + ½ m v =.66 x 10 6 14. A body of mass m and speed v has kinetic energy J. A second body of mass m/ moves at speed v. The kinetic energy of this second body is KE = ½ (m/)(v) = (½ mv ) = 4 J 15. A body moving along a straight-line has mass 3.0 kg and kinetic energy 4 J. The motion is then opposed by a net force of 4.0 N. The body will come to rest after travelling a distance of (use work-kinetic energy relationship) W = ½ mv ½ mv 1 W = - F fr d - 4d = 0 4 d = 6 m Conservation of energy law For the system that has only mechanical energy (ME = PE + KE) and no mechanism for loosing that energy (isolated system): ME 1 = ME = ME 3 = ME 4 mgh 1 + ½ mv 1 = mgh + ½ mv = 1.A pebble rolls off the roof of Science Hall and falls vertically. Just before it reaches the ground, the pebble's speed is 17 m/s. Neglect air resistance and determine the height of Science Hall. PE top = KE bottom mgh = ½ mv gh = ½ v h = v /g = 89/(x9.8) = 15 m Use the following to answer questions - 5: A 10.0-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second horizontal surface as shown in the figure..what is the kinetic energy of the crate as it slides on the upper surface? KE = ½ mv = 80 J 3. While the crate slides along the upper surface, how much gravitational potential energy does it have compared to what it would have on the lower surface? PE = mgh = 90 J 4.What is the speed of the crate when it arrives at the lower surface? mgh + ½ mv 1 = ½ mv 90 + 80 = ½ mv 370 = 5.0 v v = 8.6 m/s 5.What is the kinetic energy of the crate as it slides on the lower surface? mgh + ½ mv 1 = ½ mv = 370 J or KE = ½ mv = 370 J
9 6.A roller-coaster car is moving at 0 m/s along a straight horizontal track. What will its speed be after climbing the 15-m hill shown in the figure if friction is ignored? KE bottom = PE top ½ m v 1 = mg h + ½ m v ½ v 1 = g h + ½ v 00 = 150 + ½ v v = 10 m/s 7.A 3.0-kg block falls from rest through a distance of 6.0 m in an evacuated tube near the surface of the earth. What is its speed after it has fallen the 6.0 m distance? ΔPE = KE mg Δh = ½ mv g Δh = ½ v 10 = v v = 11 m/ s 8.A bicyclist is traveling at a speed of 0.0 m/s as he approaches the bottom of a hill. He decides to coast up the hill and stops upon reaching the top. Determine the vertical height of the hill. ½ m v = mg h ½ v = g h h = 0 m 9.A skier leaves the top of a slope with an initial speed of 5.0 m/s. Her speed at the bottom of the slope is 13 m/s. What is the height of the slope? mgh + ½ mv 1 = 0 + ½ m v gh + ½ v 1 = 0 + ½ v 10h + 1.5 = 84.5 h = 7.3 m 10.A roller coaster starts from rest at the top of an 18-m hill as shown. The car travels to the bottom of the hill and continues up the next hill that is 10.0 m high. How fast is the car moving at the top of the 10.0-m hill if friction is ignored? mgh 1 + ½ mv 1 = mgh + ½ m v gh 1 + ½ v 1 = gh + ½ v 180 + 0 = 100 + ½ v v = 13 m/s What if the friction is present? Friction converts part of kinetic energy of the object into heat energy. We say that the frictional force has dissipated energy. This energy equals to the work done by the friction and it doesn t belong to the object alone but is shared with environment. Mechanical energy is not conserved any more. ME 1 F fr d = ME (W fr = F fr d) 1. A child riding a bicycle has a total mass of 40.0 kg. The child approaches the top of a hill that is 10.0 m high and 100.0 m long at 5.0 m/s. If the force of friction between the bicycle and the hill is 0.0 N, what is the child s velocity at the bottom of the hill? (Disregard air resistance. g = 9.81 m/s.) ME 1 F fr d = ME mgh + ½ mv 1 F fr d = ½ mv v = 11 m/s. A parachutist with a mass of 50.0 kg jumps out of an airplane at an altitude of 1.00 x 10 3 m. After the parachute deploys, the parachutist lands with a velocity of 8.00 m/s. Find the energy that was lost to air resistance during this jump. (g = 9.81 m/s.) ME 1 = mgh = 490500 J ME = ½ mv = 1600 J energy lost = 490500 1600 J = 488900 J 3..An ideal spring is used to fire a 15.0-g block horizontally across a frictionless table top. The spring has a spring constant of 0 N/m and is initially compressed by 7.0 cm. The speed of the block as it leaves the spring is: Elastic potential energy of the spring is converted into kinetic energy of the block: EPE = KE
10 Power is the work done in unit time or energy converted in unit time measures how fast work is done or how quickly energy is converted. Power is a scalar quantity. Units: 1 W(Watt) = 1 J/ 1s There is another way to calculate power P = Fv 1. The kinetic energy of a car is 8 x 10 6 J as it travels along a horizontal road. How much power is required to stop the car in 10 s? W = ΔKE = - 8 x 10 6 J P = W/t = 8 x 10 5 W.An escalator is used to move 0 people (60 kg each) per minute from the first floor of a department store to the second floor, 5 m above. The power required is approximately: P = W/t = mgh/t = 0(60)(10)(5)/(60) = 1000 W 3. A 51-kg woman runs up a flight of stairs in 5.0 s. Her net upward displacement is 5.0 m. Approximately, what average power did the woman exert while she was running? P = W/t = mgh/t = (51)(9.80)(5.0)/(5.0) = 0.50 kw 4.What power is needed to lift a 49-kg person a vertical distance of 5.0 m in 0.0 s? P = W/t = mgh/t = (49)(9.80)(5.0)/(0.0) = 10 W 5.An escalator is 30.0 meters long and slants at 30.0 relative to the horizontal. If it moves at 1.00 m/s, at what rate does it do work in lifting a 50.0 kg man from the bottom to the top of the escalator? F = mg sin 30.0 0 = 45 N P = Fv = (45)(1.00) = 45 W 6.How much power is needed to lift a 75-kg student vertically upward at a constant speed of 0.33 m/s? P = Fv = mg v = (75)(9.8)(0.33) = 43 W 7.A warehouse worker uses a forklift to lift a crate of pickles on a platform to a height.75 m above the floor. The combined mass of the platform and the crate is 07 kg. If the power expended by the forklift is 1440 W, how long does it take to lift the crate?. P = W/t = mgh/t 1440 = (07)(9.8)(.75)/t t = 3.87 s 8. An electric train develops a power of 1.0 MW when travelling at a constant speed of 50 ms 1. The net resistive force acting on the train is P = Fv F = 0 kn 9. Engine X is stated to be more powerful than engine Y. Which of the following is the correct comparison of the engines? A. Engine X produces a larger force than engine Y. B. Engine X produces more useful energy than engine Y. C. Engine X produces more useful energy per unit time than engine Y. D. Engine X produces more power for a longer time than engine Y. Efficiency is the ratio of how much work, energy or power we get out of a system compared to how much is put in. W E P eff = = = W E P out out out in in in No units Efficiency can be expressed as percentage by multiplying by 100%. No real machine or system can ever be 100% efficient, because there will always be some energy changed to heat due to friction, air resistance or other causes. 1. A machine lifts an object of weight to 1.5 10 3 N a height of 10 m. The machine has an overall efficiency of 0 %. The work done by the machine in raising the object is mgh = 0. W W = 7.5 10 4 J