GEN PHYSICS 1 REVIEWER NOTES FOR FINALS-Work,Energy,Power, Momentum and Impulse, Quizzes of Physics

Download GEN PHYSICS 1 REVIEWER NOTES FOR FINALS-Work,Energy,Power, Momentum and Impulse and more Quizzes Physics in PDF only on Docsity! Welcome! Exam Pointers Topic Number of items Number of Points Work, Energy and Power 9 items 12 points Impulse and Momentum 6 items 12 points Center of mass and Collision 9 items 12 points Moment of Inertia 7 items 13 points Torque 8 items 12 points Make Meaning 1 item 9 points TOTAL 40 ITEMS 70 POINTS WORK, ENERGY AND POWER An object has a mass of 5kg. Calculate its potential energy 0.003km above the ground. GIVEN: m= 5 kg h= 0.003 km or 3.0 m GPE = mgh GPE = 5 kg (9.8 m s2 )(3.0 m) 𝑮𝑷𝑬 = 𝟏𝟒𝟕 𝑱 IDENTIFY THE FORMULA APPLY SUBSTITUTION PROPERTY SOLVE FOR THE FINAL ANSWER WORK, ENERGY AND POWER A crane lifts a 1.8 𝑥103 kg load of scrap metals to a distance of 0.015km in 20 s. How much horsepower is used in delivering the load? GIVEN: m= 1.8 𝑥103 kg h= 0.015 km or 15 m t= 20 s P = W t P = mgh t P = 1.8 x103kg ( 9.8 m s2)(15 m) 20 s P = 13, 230 watts 13, 230 watts = 13, 230 watts x 1 ℎ𝑝 746 𝑤𝑎𝑡𝑡𝑠 13, 230 watts =17.73 hp Recall: W = Fd(horizontal) or W = Fh(vertical) F = ma (horizontal) or F = mg (vertical) 𝐻𝑒𝑛𝑐𝑒, 𝑊 = 𝑚𝑔ℎ 1 Hp= 746 watts IDENTIFY THE FORMULA DERIVE THE FORMULA AND SOLVE FOR POWER APPLY SUBSTITUTION PROPERTY SOLVE FOR POWER CONVERT WATTS TO HORSEPOWER WORK, ENERGY AND POWER What power in watts is employed in lifting a 114.64 pounds mass to a height of 20m in 1 minute? GIVEN: m= 114.64 𝑝𝑜𝑢𝑛𝑑𝑠 or 52 kg h= 20 m t= 1 minute or 60 seconds P = W t P = mgh t P = 52 kg ( 9.8 m s2)(20 m) 60 s 𝑷 = 170 watts IDENTIFY THE FORMULA APPLY SUBSTITUTION PROPERTY SOLVE FOR POWER DERIVE THE FORMULA AND SOLVE FOR POWER You've been rowdy and obnoxious in a restaurant and are now in the process of being thrown out by the scruff of the neck by the guard. The guard has hold of you for 5.0 s and you are taken from a seated position to a final speed of 2.75 m/s. If your mass is 70.0 kg, what is the average force did the bouncer exert on you? Assume all motion is in a straight line. 𝐹 = 𝑚𝑎 𝐹 = 𝑚( 𝑣𝑓 − 𝑣𝑖 𝑡 ) 𝐹 = 70 𝑘𝑔 ( 2.75 𝑚 𝑠 − 0 5.0 𝑠 ) 𝐹 = 38.5 𝑁 M O M E N T U M A N D I M P U L S E GIVEN: 𝑡 = 5.0 𝑠 𝑣𝑓 = 2.75 𝑚 𝑠 𝑣𝑖 = 0 𝑚 𝑠 𝑚 = 70 𝑘𝑔 IDENTIFY THE FORMULA DERIVE THE FORMULA APPLY SUBSTITUTION PROPERTY SOLVE FOR FORCE A lion of mass 240 kg leaps at a hunter with a horizontal velocity of 24m/s. The hunter has an automatic rifle firing bullets of mass 30 g with a muzzle speed of 1260m/s and he attempts to stop the lion in midair. How many bullets would the hunter have to fire into the lion to stop its horizontal motion? Assume the bullets stick inside the lion. M O M E N T U M A N D I M P U L S E GIVEN: 𝑚1 = 240 𝑘𝑔 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑙𝑖𝑜𝑛 ; 𝑣1 = 24 𝑚 𝑠 (𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓𝑙𝑖𝑜𝑛) 𝑚2 = 30 𝑔 𝑜𝑟 0.03 𝑘𝑔 (𝑚𝑎𝑠𝑠 𝑜𝑓 𝑏𝑢𝑙𝑙𝑒𝑡𝑠); 𝑣2 = 1260 𝑚 𝑠 (𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑏𝑢𝑙𝑙𝑒𝑡𝑠) Momentum of Lion pLion = mv pLion = 240 kg (24 m s ) pLion = 5 760 kg m/s Momentum of Bullets pbullets = mv pbullets = 0.03 kg (1 260 m s ) pbullets = 37.8 kg m/s Number of bullets = plion pbullets = 5 760 kg m s 37.8 kg m s = 152 bullets C O L L I S I O N Mass A moving at 4.5 m/s hits Mass B which is initially at rest. Mass A is 5.2kg while Mass B is 2.7kg. What is the initial momentum of Mass A? COLLISION GIVEN: Mass of A= 5.2 kg Speed of A= 4.5 m/s pinitial = m1v1 pinitial = (5.2 kg)(4.5 m s ) 𝒑𝒊𝒏𝒊𝒕𝒊𝒂𝒍 = 𝟐𝟑. 𝟒 𝒌𝒈 𝒎 𝒔 IDENTIFY THE FORMULA APPLY SUBSTITUTION PROPERTY SOLVE FOR MOMENTUM M O M E N T O F I N E R T I A MOMENT OF INERTIA A 2-kg rectangular thin plate with a length of 0.5 m and width of 0.2 m. The axis of rotation is located at the center of the rectangular plate. What is the radius of gyration? GIVEN: m= 2 kg l= 0.5 m w= 0.2 m k= ? 𝑘 = 𝐼 𝑚 𝑘 = 0.048ഥ3 kg m2 2 𝑘𝑔 𝑘 = 0.0241ത6𝑚2 𝑘 = 0.16 𝑚 I = 1 12 M a2 + b2 I = 1 12 2 kg [ 0.5m 2+ 0.2m 2] 𝑰 = 𝟎. 𝟎𝟒𝟖ഥ𝟑 𝐤𝐠 𝐦𝟐 T O R Q U E A 7.5 N force is applied as shown in the figure below. Calculate the torque supplied by the wrench. Given: F= 7.5 N r= 16 cm or 0.16 m Ɵ= 47˚ T O R Q U E τ = 𝐹𝑟𝑠𝑖𝑛Ɵ τ = (7.5 𝑁)(0.16𝑚)(𝑠𝑖𝑛47) τ = −0.88 𝑁𝑚 It moves in a clockwise direction Find the distance if 90 N force acts at the end of a 12-cm wrench as shown. r= 12 cm 1 𝑚 = 100 𝑐𝑚 12 𝑐𝑚 = 12 𝑐𝑚 𝑥 1 𝑚 100 𝑐𝑚 12 𝑐𝑚 = 0.12 𝑚 T O R Q U E Given: F= 90 N r= 12 cm Ɵ= 65˚