1. Brizah kicked a soccer ball weighing 8 kilograms and it moves 20m/4s2. Suppose it covers a 10-meter distance. How much work is done?Given: m=8 kg; a=20m/4s2 or 5m/s2; d=10 mRequired: WorkSolution:A. F= amB. W= Fd = 5m/s2 (8 kg) = 40 N (10m) = 40 kg m/s2 W = 400 JoulesF= 40 N2. Assume that the soccer ball weighted 10 000 grams and moves at 2.5m/s2. With the same amount of work done, how much distance does it covers?Given: m= 10 000 grams or 10 000 g x 1 kg = 10 kg; a= 2.5m/s2; w= 400 Joules 1000Required: distanceSolution:A. F=amB. W= Fd= 2.5m/s2 (10 kg) d= W F= 25 N F = 400 J 25 N d= 16 meters3. The cart was pulled at a distance of 25 m with 650 Joules. Find the Force exerted in the cart.Given: w= 400 Joules; d= 25 mRequired: forceSolution:W= Fd F= W D= 650 25 F= 26 N4. If the mass of a push cart is 46 kg and moves at a speed of 0.4 m/s. What is the KE? Given: m= 46 kg; v=0.4 m/sRequired: KESolution:KE= (46) (0.4 m/s)2= (46) (0.16) KE= 3.68 Joules5. A boy pushed a 3-N box weighing 40 kilograms at a distance of 15 meters. What is its velocity?Given: F= 675 N; m= 40 kg; d= 15 metersRequired: velocitySolution: A. W= Fd = 3 N (15 m)W= 45 JB. KE= mv2=2(45)/ 40 mv2= KE=90/40 V2= 2(KE)/m=2.25 V=2(KE)/m V=1.5 m/s6. With 2.5J of Kinetic energy, find the mass of the toy car carrying a 12 kg mass running at a speed of 0.5m/s2.Given: KE= 2.5 J; m2= 12 kg; v= 0.5m/sRequired: massSolution:KE= mv2Tm=m2+mo mv2 = KEmo= Tm- m2 M = 2(KE)/v2 = 20- 12 = 2(2.5 J)/ (0.5m/s)2mo = 8 kg = 5/0.25 m = 20 kg

7. A canon ball was pitched vertically on a height of 0.55 km. If the Potential Energy is 379 500 J, what is its mass?Given: PE= 379 500 J; h= 0.55 kmRequired: massSolution:PE=mgh mgh= PE m= PE/mgh= 379 500/ (10m/s2)= 379 500/ 5500 m= 69 kg8. Given the force 56N and the height of 0.0055 km, find the PE.Given: F= 56 N; h= 0.0055 km or 5.5mRequired: PESolution:PE=mgh= 56 N (5.5 m) PE= 308 Joules9. Calculate the height. A girl who weighed 70 kg jumped on a trampoline with a PE of 2646 Joules.Given: m= 70 kg; PE= 2 646 JRequired: heightSolution:PE= mghH = PE/mg = 2 646/ (70) (10) = 2 646/ 700 h= 3.78 meters10. A book weighing 0.2 kg fell down from a 160 m shelf in a distance of 40 m. What is its KE?Given: m= 0.2 kg; hf = 160 m; hi= 40 mRequired: KESolution:A. PE= mgh= 0.2 kg (10m/s2) (160) PE= 320 JoulesB. KE= 0PE= mg (hf- hi)= 0.2 kg (10m/s2) (160-40)= 0.2 kg (10m/s2) (120m) PE= 240 JoulesC. KE+PE = 320 J KE = 320- 240 J KE = 80 J11. The cat jumped from a 20 m building, as soon as it reaches the ground, its Kinetic Energy is 400 Joules. (a.) What is the mass of the cat? (b.) What is its P.E. if it fell halfway to the ground?Given: h= 20 meters; K.E. = 400 JoulesRequired: (a.) mass; (b.) P.E.Solution:A. PE=KEPE=mgh m=PE/gh = 400J/ (10 m/s2) (20 m) = 400kg m2/s2/ 200m2/s2 m= 2 kilogramsB. PE=mgh = 2kg (10 m/s2) (20 m/ 2) = 20 kg m/s2 (10 m)PE= 200 Joules

12. Suppose an elevator carries 4 people, weighing 22 kg each with a time interval of 4 minutes. How much power does it possess in watts if it has to stop on the second floor on a 60-m, 3 story building.Given: m= 4 (22 kg)=88 kg; t= 4 minutes x 60 seconds = 240 seconds; h= 60 meters 1 minuteRequired: PowerSolution:A. W= FdB. P= W/t = mgh = 35 200 J/ 240 sec = 88 kg (10 m/s2) (40 m) P= 146. 67 WattsW= 35 200 Joules13. Izah is pulling a sack for 2 minutes. Find the work done with the power of 45 Watts.Given: t= 2 minutes x 60 seconds = 120 seconds; P= 45 watts 1 minuteRequired: WorkSolution: P =W/tW= Pt = 45 watts (120 seconds)W= 5 400 Joules14. How does it take to pull a sack weighing 30 N in a distance of 25 meters and a power of 44 760 hp?Given: F= 30 Newton; d= 25 meters; P= 44 760 hp x 1 hp = 60 watts ? 746 wattsRequired: timeSolution:P= W/tP= Fd/tPt= Fd t = Fd/P = 30 N (25 m) / 60 W = 750 J/ 60 W t = 12.5 seconds

15. How much input energy does the machine have when the energy output is 36 and the Efficiency is 90Given: Energy Output= 36; Efficiency= 90Required: Input EnergySolution:E = Output Energy x 100 Input Energy 90 = 36 x 100 I.E.Input Energy = 36 0.9Input Energy = 40 16. A coin was dropped to the ground from a height of 0.5 meters and a force of 1.9 N. The specific heat of the coin is 0.44 J/g C and the change of temperature is 0.50C. Calculate its mass in grams.Given: h= 0.5m; F= 1.9N; c= 0.44 J/g C; T= 0.50C Required: massSolution:W=HH=mcT m= H / cT = Fd / cT = 0.5m (1.9 N ) / 0.44 J/g C (0.50C ) = 0.95 J / 0.22 J/gm= 4.32 grams17. If the same coin was dropped to the pool of water from a height of 2 meters and a force of 3.5 Newton. Find the change in temperature.Given: h= 2 m; F= 3.5 N; m= 4.32 grams; cwater= 4.2 J/g C Required: T Solution:W=HH=mcT T = H / c (m) = Fd / c (m)

= 2 m (3.5 N) / 4.2 J/g C (4.32 g) = 7 J / 4.2 J/g C (2.2 g) = 7 J / 9.24 JC T = 0.76 C 18. Find the amount of heat of water if a 0.003-kg marble was dropped into a pool. The change in temperature is 0.6 CGiven: m= 0.003 kg x 1000 grsms = 3 g; cwater=4.2 J/g C; T=0.6 C1 kgRequired: HSolution:H= mcT = 3 g (4.2 J/g C) (0.6 C) H= 7.56 Joules