How much work does a 154 lb. student do when climbing a flight of stairs that are 6 meters in height and 30 meters in length? If the stairs are climbed.

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How much work does a 154 lb. student do when climbing a flight of stairs that are 6 meters in height and 30 meters in length? If the stairs are climbed in 4 seconds calculate the power in Watts and horsepower. (1kg = 2.2 lbs.) Mass in kg 154 lb x 1 kg = 70 kg 2.2 lb. Weight of student Fg = mg Fg = (70 kg) (9.8 m/s2) = 686 N W = Fd W = (686 N) (6 m) = 4,116 N P = W P = 4,116 N = 1,029 watts (1HP = 746 watts) t 4 s 1029 watts x 1 HP = 1.4 HP 746 watts

Conversion factors How many inches are there in 2 kilometers? (1 mile = 1.61 km) (1 mile = 5280 ft.) 2 km x 1 mile x 5280 feet x 12 inches = 126,720 1.61 km 1 mile 1 foot 1.61 = 78,708 feet

What is“energy” Two forms of mechanical energy are? Potential energy – gravitational potential energy “energy of position” Kinetic energy – “energy of motion”

Kinetic Energy – energy of motion When chasing prey, a cheetah can reach a top speed of 120 m/s for a short period of time. What is the kinetic energy of a 40 kg cheetah when moving at this speed? KE = ½ mv2 KE = ½ (40 Kg) (120 m/s)2 KE = 288,000 Joules

Kinetic Energy KE = ½ mv2 Why do stopping distances increase by four times when the speed is doubled? Calculate the kinetic energy of a car with a total mass of 900 kg going 40 m/s? …… 80 m/s? For car going 40 m/s KE = ½ (900 kg) (40m/s)2 = 720,000 J For car going 80 m/s KE = ½ (900 kg) (80m/s)2 = 2,880,000 J So stopping distances increase by four times when the speed is doubled because…….. – there is four times the kinetic energy to overcome by the braking force. Work energy theorem – whenever work is done energy changes work = ∆KE so Fd = ∆KE the force applied by the brakes times the stopping distance equals the change in kinetic energy (720,000 J or 2,880,000 J)

The amount of potential energy of an object is equal to the amount of work done to get an object to that position. An 8 kg rock is carried to the top of a 5m hill. What is the potential energy of the rock? PEg = mgh F PEg = (8 kg) (9.8 m/s2) (5m) PEg = 392 Joules Amount of work done to get rock to the top of the hill? 392 J Amount of kinetic energy of rock just before it hits the ground? 392 J

Conservation of Energy The law of conservation of energy states that energy cannot be created or destroyed. Energy can be transformed from one form into another but the total amount never changes.

PEg = mgh For a ball moving down a ramp, in the absence of friction, the gravitational potential energy at the top of the ramp would be equal to the kinetic energy at the bottom of the ramp. KE = 1/2 mv2 6m A 2kg ball rolls down a frictionless ramp. The ball is 6 vertical meters above a level section of track. What is the velocity of the ball as it moves on the level section of the track? If there is no friction then PEg (top) = KE (bottom) So mgh = 1/2 mv2 Then simplify to gh = 1/2 v2 Rearrange to v2 = 2gh So v = √2gh So v = √ (2) (9.8 m/s2) (6 m) = 10.8 m/s