New version page

TAMU PHYS 1401 - Work & CoE solutions

This preview shows page 1-2-3-4-5 out of 15 pages.

View Full Document
View Full Document

End of preview. Want to read all 15 pages?

Upload your study docs or become a GradeBuddy member to access this document.

View Full Document
Unformatted text preview:

jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) 1This print-out should have 40 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsThe Joule and the kilowatt-hour are bothunits of energy.19.2 kW · h is equivalent to how manyJoules?Correct answer: 6.912 × 107J.Explanation:1 W = 1 J/s19.2 kW · h = (19.2 kW · h) ·1000 WkW·3600 s1 h= 6.912 × 107W · s=6.912 × 107J .002 10.0 pointsA cheerleader lifts his 59.9 kg partner straightup off the ground a distance of 0.557 m beforereleasing her.The acceleration of gravity is 9.8 m/s2.If he does this 27 times, how much work hashe done?Correct answer: 8828.19 J.Explanation:The work done in lifting the cheerleaderonce isW1= m g h= (59.9 kg)(9.8 m/s2)(0.557 m)= 326.97 J .The work required to lift her n = 27 times isW = n W1= (27)(326.9 7 J) = 8828.19 J.003 10.0 pointsA student weighing 709 N climbs at constantspeed to the top of an 9 m vertical rope in14 s.What is the average power expended by thestudent to overcome gravity?Correct answer: 455.786 W.Explanation:Let : F = 709 N ,d = 9 m , andt = 14 s .The power expended isP =Wt=F dt=(709 N) (9 m)14 s= 455.786 W .004 10.0 pointsYou leave your 75 W portable color TV on for7 hours each day and you do not pay attentionto the cost of electricity.If the dorm (or your parents) charged youfor your electricity use and the cost was$0.1 /kW · h, what would be your monthly(30 day) bill?Correct answer: 1.575 dollars.Explanation:Let : P = 75 W andt = 7 h/day ,The energy consumed in each day isW = P t= (75 W) (7 h/day) ·kW1000 W= 0.525 kW · h/day.In 30 days, you would use(30 day) (0.525 kW · h/day) = 15.75 kW · h,which would cost you(15.75 kW · h) ($0.1/kW · h) =$1.575 .jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) 2005 10.0 pointsA 105 kg physics professor has fall en intothe Grand Canyon. Luckily, he managedto grab a branch and is now hanging 87 mbelow the r im. A student (majoring in lin-guistics and physics) decides to perform arescue/experi ment using a nearby horse. Af-ter lowering a rope to her fall en hero andattaching the other end to the horse, the stu-dent measures how long it takes for the horseto pull the fallen physicist to the rim of theGrand Canyon.The acceleration of gravity is 9.8 m/s2.If the horse’s output power is truly 1 horse-power (746 W), and no energy is lost to fric-tion, how long should the process take?Correct answer: 120.004 s.Explanation:The work agai nst gravity which is necessaryto raise the professor to t he rim of the canyonis given byW = m g ∆h= (105 kg) (9.8 m / s2) (87 m)= 89523 J .If the horse’s power output is one horsepower,then the the time for the horse to provide thenecessary work is given bytup=WPoutput=89523 J746 W= 120.004 s .006 10.0 pointsPotential energy and kinetic energy are formsof what kind of energy?1. chemical2. nuclear3. mechanical correct4. electromagnetic5. heatExplanation:007 10.0 pointsA Joule is a measure of1. density.2. distance.3. volume.4. momentum.5. energy. correctExplanation:008 10.0 pointsShawn and his bike have a t otal mass of58.7 kg. Shawn rides his bike 0.77 km i n13.9 min at a constant velocity.The acceleration of gravity is 9.8 m/s2.What is Shawn’s kinetic energy?Correct answer: 25.0184 J.Explanation:Shawn’s velocity isv =dt.Thus his kinetic energy isK =12mv2=12(58.7 kg)0.77 km13.9 min·1000 m1 km·1 min60 s2= 25.0184 J .009 10.0 pointsTim, with mass 71.6 k g, climbs a gymnasiumrope a distance of 2 m.The acceleration of gravity is 9.8 m/s2.How much potential energy does Tim gain?Correct answer: 1403.36 J.jonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) 3Explanation:Let : m = 71.6 kg ,h = 2 m , andg = 9.8 m/s2.Potential energy isU = m g h= (71.6 kg) (9.8 m/s2) (2 m)=1403.36 J .010 (part 1 of 3) 10.0 pointsA 40.0 kg child is in a swing that is attachedto ropes 2.30 m long.The acceleration of gravity is 9.81 m/s2.Find the gravitational pot ential energy as-sociated with the child relative to the child’slowest position under the following condi-tions:a) when the ropes are horizontal.Correct answer: 902.52 J.Explanation:Basic Concept:Ug= mghGiven:ℓ = 2.30 mm = 40.0 kgg = 9.81 m/s2Solution:The child is at a height h1= ℓ from thelowest point, soUg= mgh1= (40 kg)(9.81 m/s2)(2.3 m)= 902.52 J011 (part 2 of 3) 10.0 pointsb) when the ro pes make a 36.0◦angle withthe vertical.Correct answer: 172.366 J.Explanation:Given:θ = 36.0◦Solution:The child is a distance ℓ cos θ from the at-tachment point, so h2= ℓ − ℓ cos θ andUg= mgh2= mgℓ(1 − cos θ)= (40 kg)(9.81 m/s2)(2.3 m) · (1 − cos 36◦)= 172.366 J012 (part 3 of 3) 10.0 pointsc) at the bottom of the circular arc.Correct answer: 0 J.Explanation:Solution: The child is a t the lowest point,so h3= 0, andUg= mgh3= (40 kg)(9.81 m/s2)(0 m)= 0 J013 10.0 pointsA block of mass m sli des on a horizontalfrictionless table with an initial speed v0. Itthen compresses a spring of force constant kand is brought to rest.vmkmµ = 0How much is the spring compressed x fromits natural length?1. x = v0m kg2. x = v0rmkcorrect3. x =v202 g4. x = v0kg mjonnalagadda (saj2436) – Work & CoE – ayida – (4355-3) 45. x = v0rkm6. x = v0m gk7. x = v0mk g8. x = v0skm g9. x = v0rm gk10. x =v202 mExplanation:Tota l energy is conserved (no friction). Thespring is compressed by a dist ance x from itsnatural length, so12m v20= Ei= Ef=12k x2, orx2=mkv20, thereforex = v0rmk.Anyone who checks to see if the units ar ecorrect should get this problem correct.014 10.0 pointsA bead slides without friction around a loop-the-loop. The bead is released from a heightof 11.6 m from the bottom of the loop-the-loop which has a radius 3 m.The acceleration of gravity is 9.8 m/s2.11.6 m3 mAWhat is its sp eed at point A ?Correct answer: 10.4766 m/s.Explanation:Let : R = 3 m andh = 11.6 m .From conservation of energy, we haveKi+ Ui= Kf+ Uf0 + m g h =m v22+ m g (2 R)v2= 2 g (h − 2 R) .Thereforev =p2 g (h − 2 R)=r2 (9.8 m/s2)h11.6 m − 2 (3 m)i= 10.4766 m/s .015 (part 1 of 3) 10.0 pointsA block starts at rest and slides down a fric-tionless track. It leaves the tra ck horizontally,flies through the air, and subsequent ly strikesthe ground.bbbbbbbbbbbb570 g4.9 m2.4 mxvWhat is the speed of the ball when it leavesthe t rack? The acceleration of g r avity is9.81 m/s2.Correct answer: 7.00357 m/s.Explanation:Let : g = −9.81 m/s2,m = 570 g , andh1= 2.5 m .jonnalagadda (saj2436) – Work


View Full Document
Loading Unlocking...
Login

Join to view Work & CoE solutions and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Work & CoE solutions and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?