Total Energy before = Total Energy After Conservation of Energy Total Energy before = Total Energy After Comes from = Goes to Assets = Expenditures Fd + mgh + 1/2mv2 = Fd + mgh + 1/2mv2

Fd + mgh + 1/2mv2 = Fd + mgh + 1/2mv2 A 250 kg cart going 4.5 m/s rolls down a 1.75 m tall hill. What is the velocity of the cart at the bottom? v = 4.5 m/s 250 kg 1.75 m Fd + mgh + 1/2mv2 = Fd + mgh + 1/2mv2 7.4 m/s

Example 1 An 890 kg cart rolling 6.2 m/s along a level surface hits a 3.6 m long puddle that exerts 3200 N of average retarding force. What is the cart’s velocity after this? v = 6.2 m/s 890 kg 3.6 m (Puddle - Exerts 3200 N of retarding force) 3.5 m/s

Example 2 A 350 kg cart is going 4.6 m/s. For what distance must a person exert a forward force of 53 N so that when the cart gets to the top of a 1.8 m tall hill it is going 2.4 m/s? v = 4.6 m/s 350 kg 1.8 m 66 m

Example 3 A 0.124 kg pine cone falls 45.0 m from a tree. It is going 22.0 m/s when it strikes the ground. What is the average force of air friction that acts on the pine cone as it falls? 0.550 N

(15 kg)(9.81 N/kg)(2.15 m) = 1/2(15 kg)v2 u = 0 What speed at the bottom? h = 2.15 m (15 kg)(9.81 N/kg)(2.15 m) = 1/2(15 kg)v2 v = 6.4948… 6.5 m/s

(15 kg)(9.81 N/kg)(2.15 m) + 1/2(15 kg)(5.8 m/s)2= 1/2(15 kg)v2 u = 5.8 m/s What speed at the bottom? h = 2.15 m (15 kg)(9.81 N/kg)(2.15 m) + 1/2(15 kg)(5.8 m/s)2= 1/2(15 kg)v2 v = 8.7076… 8.7 m/s

What speed at the top? 15 kg h = 4.25 m h = 2.15 m u = 8.6 m/s (15 kg)(9.81 N/kg)(2.15 m) + 1/2(15 kg)(8.6 m/s)2 =(15 kg)(9.81 N/kg)(4.25 m) 1/2(15 kg)v2 v = 5.723… 5.7 m/s

v = 3.68 m/s 2.34 kg The hammer pushes in the nail 3.50 mm. (.00350 m). What force did it exert on the nail 1/2(2.34 kg)(3.68 m/s)2 = F(.0035 m) 4530 N

(125 kg)(9.81 N/kg)(18.0 m + 1.50 m) = F(1.50 m) A 125 kg experiment falls 18.0 m and has its velocity arrested in 1.50 m by an air bag. What force does the air bag exert on the experiment in stopping the experiment? 18 m (125 kg)(9.81 N/kg)(18.0 m + 1.50 m) = F(1.50 m) F = 15,941.25 15,900 N

(450 kg)(9.81 N/kg)(1.75 m) + 1/2(450 kg)(5.8 m/s)2 = 1/2(15000 N/m)x2 What distance will it compress the 15000 N/m spring in stopping the cart? 450 kg u = 5.8 m/s h = 1.75 m (450 kg)(9.81 N/kg)(1.75 m) + 1/2(450 kg)(5.8 m/s)2 = 1/2(15000 N/m)x2 x = 1.428… 1.4 m

What must be the spring constant to give the 115 g marble a velocity of 2.13 m/s on top of the hill if the spring is compressed 3.15 cm? h = .452 m 1/2k(.0315 m)2 = 1/2(.115 kg)(2.13 m/s)2 + (.115 kg)(9.81 N/kg)(.452 m) k = 1553.63… 1550 N/m