Work and Energy Dr. Walker

Work Work is the product of the force on an object and the distance through which the object is moved: the quantity force × distance When we lift an object against gravity, we perform work Work is increased when We lift something heavier (F = ma) We accelerate an object at a higher rate (F = ma) We move something farther (W = Fd)

Work Units The unit of measurement for work combines a unit of force, N, with a unit of distance, m. The unit of work is the newton-meter (N•m), also called the joule. Remember, W = Fd One joule (J) of work is done when a force of 1 N is exerted over a distance of 1 m (lifting an apple over your head).

Examples of Work An archer stretches her bowstring, doing work against the elastic forces of the bow. When the ram of a pile driver is raised, work is required to raise the ram against the force of gravity. When you do push-ups, you do work against your own weight. Bringing an automobile up to speed or in slowing it down involves work.

Example 1 Suppose that you apply a 60-N horizontal force to a 32-kg package, which pushes it 4 meters across a mailroom floor. How much work do you do on the package?

Example 1 Suppose that you apply a 60-N horizontal force to a 32-kg package, which pushes it 4 meters across a mailroom floor. How much work do you do on the package? W = Fd = 60 N x 4 m = 240 J Notice we didn’t need the mass (tricksy hobbitses…)

Example 2 A 6600 kg building is moved a distance of 2.5 m. What is the amount of work performed?

Example 2 A 6600 kg building is moved a distance of 2.5 m. What is the amount of work performed? What is the force of the building? F = ma = (6600 kg)(9.8 m/s2) = 6.5 x 104 N What is the work? W = Fd = (6.5 x 104 N)(2.5 m) = 1.6 x 105 J

Example 3 A lifter performs a 440 lb (200 kg) deadlift. The distance from the floor to lockout is 0.6 m. How much work is performed?

Example 3 A lifter performs a 440 lb (200 kg) deadlift. The distance from the floor to lockout is 0.6 m. How much work is performed? W = Fd W = (ma)d W = (200 kg)(9.8 m/s2)(0.6 m) = 1179 J or 1.18 kJ

Energy The ability to produce change within a system Potential Energy (“position”) Elastic potential energy (an extended rubber band) Chemical energy (stored in chemical bonds, released during reactions) Gravitational potential energy (released during a fall) Kinetic Energy (“motion)

Gravitational Potential Energy The amount of gravitational potential energy possessed by an elevated object is equal to the work done against gravity to lift it. The upward force of a stationary object is mass x acceleration due to gravity W = Fd As a result, for gravitational potential energy, it’s PE = mgh or mass x gravity x height

Example 1 What is the potential energy of a 10 kg boulder at a height of 2 m?

Example 1 What is the potential energy of a 10 kg boulder at a height of 2 m? PE = mgh PE = (10 kg)(9.8 m/s2)(2 m) = 196 J

Gravitational Potential Energy The 10 kg boulder has 200 J of potential energy at a height of 2 m. The method of getting the boulder on the platform is irrelevant.

Example 2 A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of the loaded cart is 3.0 kg and the height of the seat top is 0.45 meters, then what is the potential energy of the loaded cart at the height of the seat-top?

Example 2 A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of the loaded cart is 3.0 kg and the height of the seat top is 0.45 meters, then what is the potential energy of the loaded cart at the height of the seat-top? PE = mgh PE = (3.0 kg)(9.8 m/s2)(0.45 m) = 13.2 J Note: If you were asked for the work performed, you would’ve done the same calculation since mg = F and the height is the distance

Kinetic Energy Energy of Motion KE = ½ mv2 If the velocity is doubled, the KE increases by a factor of 4 This can be derived W = Fd, F = ma, and Vf = Vi + 2ad

Kinetic Energy Example 1 Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s.

Kinetic Energy Example 1 Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s. KE = ½ mv2 KE = ½(625 kg)(18.3 m/s)2 = 105000 J = 105 kJ

Kinetic Energy Example 2 Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a kinetic energy of 12 000 J just prior to hitting the bucket of water. If Missy's mass is 40 kg, then what is her speed?

Kinetic Energy Example 2 Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a kinetic energy of 12 000 J just prior to hitting the bucket of water. If Missy's mass is 40 kg, then what is her speed? KE = ½ mv2 12000 J = ½ (40 kg)v2 V = 24.5 m/s

Review Exercise If Missy was moving at 24.5 m/s when hitting the water, at what height was her platform?

Review Exercise If Missy was moving at 24.5 m/s when hitting the water, at what height was her platform? V = at 24.5 m/s = (9.8 m/s2)t t = 2.5 s D = 1/2gt2 D = ½ (9.8 m/s2)(2.5 s)2 D = 30.6 m Or you could have used Vf2 = Vi2 + 2ad and gotten the same answer.

Work-Energy Theorem Work = ∆KE The work is based on net work (yep, it’s a vector), which is based on the net force Net force matters when two forces oppose each other, such as when friction is present Work = DKE Fd = ½mv2 Note: This is the “maximum” speed of a system. In free fall, this is the velocity just prior to impact.

Conservation Of Energy The energy of an isolated system is constant over time Energy can be converted to other forms (i.e., potential to kinetic) It cannot be created or destroyed Not all energy is instantly “converted” from PE to KE. Total energy = KE + PE. The numbers below are “maximum” numbers.

Back To Missy… Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a kinetic energy of 12 000 J just prior to hitting the bucket of water. If Missy's mass is 40 kg, then how high was her platform? Note: Yes, we already figured this out….but how can we determine it from this information since we don’t have her speed here?

Back To Missy… Missy Diwater, the former platform diver for the Ringling Brother's Circus, had a kinetic energy of 12,000 J just prior to hitting the bucket of water. If Missy's mass is 40 kg, then how high was her platform? Through the conservation of energy, KE = PE Since she has 12000 J of KE at entry, she had 12000 J PE at the top of the platform PE = mgh 12000 J = (40 kg)(9.8 m/s2)h h = 30.6 m (and yes, it should be the same answer!!)

Putting It Together An 80 kg student climbed onto a 5 m platform. How work did the student perform to get on the platform? How much potential energy does the student have on top of the platform? If the student jumps off the platform, what velocity will the student have when they land? How long did it take the student to land?

Putting It Together An 80 kg student climbed onto a 5.0 m platform. How work did the student perform to get on the platform? W = Fd = (ma)d W = (80 kg)(9.8 m/s2)(5.0 m) = 3920 J = 3.92 kJ

Putting It Together How much potential energy does the student have on top of the platform? Remember, Work = PE = KE Short Version – Since the Work was 3920 J, PE = 3920 J Long Version PE = mgh PE = (80 kg)(9.8 m/s2)(5 .0 m) = 3920 J Notice it’s the same number!!!

Putting It Together If the student jumps off the platform, what velocity will the student have when they land? Remember PE = KE If PE = 3920 J, then KE = 3920 KE = 3920 J = ½ mv2 3920 J = ½ (80 kg)v2 V = 9.98 m/s

Putting It Together – Review How long the student was in the air after jumping off of the 5.0 m platform? One Way d = (1/2)gt2 5.0 m = ½(9.8 m/s2)t2 t = 1.01 s The Other Way Their final velocity was 9.98 m/s V = at 9.98 m/s = (9.8 m/s2)(t) t = 1.01 s (it’s the same – and should be!!)

Energy Conversion A 0.1 kg ball is taken to the top of a treehouse 20 m high. What is the potential energy of this ball? What is the kinetic energy and potential energy when it has fallen halfway? How fast was the ball moving right before impact?

Energy Conversion A 0.1 kg ball is taken to the top of a treehouse 20 m high. What is the potential energy of this ball? PE = mgh PE = (0.1 kg)(9.8 m/s2)(20 m) PE = 19.6 J

Energy Conversion A 0.1 kg ball is taken to the top of a treehouse 20 m high. What is the kinetic energy and potential energy when it has fallen halfway? PE = mgh, halfway is 10 m PE = (0.1 kg)(9.8 m/s2)(10 m) = 9.8 J Note: Since the total energy of the system is 19.6 J, the ball has 9.8 J of PE and 9.8 J of KE

Energy Conversions A 0.1 kg ball is taken to the top of a treehouse 20 m high. How fast was the ball moving right before impact? At this point, all of the energy has been converted to KE KE = (1/2)mv2 19.6 J = (1/2)(0.1 kg)v2 v = 19.8 m/s Note – Since gravity accelerates an object at roughly 10 m/s/s, these numbers should make sense. If we plugged this into our old kinematics equations, the height and final speed would check out.

Power Power is the rate of doing work (or the rate of energy transfer) The faster work is performed, the more power a system has Power is measured in watts (W) Example: Your average light bulb is 40 or 60 KW

Power One non-metric unit of power is horsepower (hp) The power of a car engine is usually measured in horsepower 1 hp = 0.75 W

Example 1 Our lifter that performed a 440 lb (200 kg) deadlift took 4.0 s to complete his rep. The distance from the floor to lockout is 0.6 m. How much power was generated?

Example 1 Our lifter that performed a 440 lb (200 kg) deadlift took 4.0 s to complete his rep. The distance from the floor to lockout is 0.6 m. How much power was generated? Power = Work/Time Power = Fd/Dt = (ma)d/Dt Power = (200 kg)(9.8 m/s2)(0.6 m)/(4.0 s) = 817 watts

Example 2 Earlier that day, our same lifter performed a 220 lb (100 kg) power clean, which took 1.0 s to complete his rep. The distance from the floor to lockout is 1.25 m (it’s a longer range of motion!). How much power was generated?

Example 2 Earlier that day, our same lifter performed a 220 lb (100 kg) power clearn, which took 1.0 s to complete his rep. The distance from the floor to lockout is 1.25 m (it’s a longer range of motion!). How much power was generated? Power = Work/Time Power = Fd/Dt = (ma)d/Dt Power = (100 kg)(9.8 m/s2)(1.25 m)/(1.0 s) = 1225 watts or 1.2 kW Notice, less time = more POWER!

Example 3 Ryan hauls his piano up two flights of stairs (22 m) in 5 minutes. The piano’s mass is 90.7 kg. What power is used?

Example 3 Ryan hauls his piano up two flights of stairs (22 m) in 5 minutes. The piano’s mass is 90.7 kg. What power is used? Power = W/T = Fd/time = (ma)d/t Power = (90.7 kg)(9.8 m/s2)(22 m)/(300 s) = 65.2 W

Example 4 What if Ryan managed to move the same 90.7 kg piano 22 m in only 20 s?

Example 4 What if Ryan managed to move the same 90.7 kg piano 22 m in only 20 s? Power = W/T = Fd/time = (ma)d/t Power = (90.7 kg)(9.8 m/s2)(22 m)/(20 s) = 978 W Less time, more POWER!!

Example 5 An engine moves a boat through the water at a constant speed of 15 m/s. The engine must exert a force of 6.0x103 N to balance the force that water exerts against the hull. What power does the engine develop?

Example 5 An engine moves a boat through the water at a constant speed of 15 m/s. The engine must exert a force of 6.0x103 N to balance the force that water exerts against the hull. What power does the engine develop? Power = Fd/t Remember, d/t = v, so Power = Fv Power = (6.0x103N)(15 m/s) = 9.0 x 104 W or 90 kW