1. PHY 231 1 PHYSICS 231 Lecture 14: revision Remco Zegers Walk-in hour: Monday 9:15-10:15 am Helproom BPS 1248 missing ID’s 129020 151313 130786 128820 152180 152183

2. PHY 231 2 Chapter 4: Newton’s Laws  First Law: If the net force exerted on an object is zero the object continues in its original state of motion; if it was at rest, it remains at rest. If it was moving with a certain velocity, it will keep on moving with the same velocity.  Second Law: The acceleration of an object is proportional to the net force acting on it, and inversely proportional to its mass: F=ma  If two objects interact, the force exerted by the first object on the second is equal but opposite in direction to the force exerted by the second object on the first: F 12 =-F 21

3. PHY 231 3 Important equations Newton’s second law:  F=ma Gravitational Force: F=mg F=Gm 1 m 2 /r 2 Earth: m 1 =m earth r=r earth g=Gm 2 /r 2 =9.8 m/s 2 Equilibrium:  F x =0  F y =0 (object not moving)  F L =0  F // =0 normal force: force perpendicular to surface the object is resting on, and balancing the component of the gravitational force perpendicular to the surface friction: F=  s n (  s : coef. of static friction) F=  k n (  k : coef. of kinetic friction)  s >  k Hooke’s law: F s =-kx (k: spring constant)

4. PHY 231 4 example 1 To lift a patient, 4 nurses grip the sheet on which the patient is lying and lift upward. If each nurse exerts an upward force of 240 N and the patient has an upward acceleration of 0.504 m/s 2 what is the weight of the patient?  F=ma (only in vertical direction) 4F nurse -F g,patient =m patient a patient 960-m patient g=0.504m patient m patient =960/(0.504+9.81)=93.1 kg w patient =93.1*9.81=913 N

5. PHY 231 5 example 2 A 5.0 kg bucket of water is raised from a well by a rope. If the upward acceleration of the bucket is 3.0 m/s 2, find the force exerted by the rope on the bucket. T FgFg  F=ma (vertical direction only) T-mg=ma T=m(g+a)=5(3.0+9.8)=64 N

6. PHY 231 6 example 3 A 1000-kg car is pulling a 300 kg trailer. Their acceleration is 2.15 m/s 2. Ignoring friction, find: a)the net force on the car b)the net force on the trailer c)the net force exerted by the trailer on the car d)the resultant force exerted by the car on the road 1000 300 F engine F tc =-F ct F engine =m total a=1300*2.15=2795 N F ct =m trailer *2.15=645 N, so F tc =-645 N a)  F car =2795-645=2150 F car =m car *2.15=2150 b)  F trailer =F tc =645 N c)F tc =-645 N 2150 mg=1000*9.8 d)F total =  (2150 2 +(-9800) 2 = 1E+04 N

7. PHY 231 7 example 4 0.5 kg A q=20 o Is there a value for the static friction of surface A for which these masses do not slide? If so, what is it? 2 kg mass:  F=ma (only vertical) T-mg=ma T-0.5g=2a 1 kg mass:  F=ma (parallel to the slope) -F g// -T+F friction =ma -mgsin(q)-T+  s mgcos(q)=ma -3.35-T+9.2  s =a No sliding: a=0, so T=0.5g=4.9 (from 0.5kg mass equation) -3.35-4.9+9.2  s =0  s =0.9 1 kg nF friction FgFg T T FgFg

8. PHY 231 8 example 5 F A force F (10N) is exerted on the red block (1 kg). The coef. of kinetic friction between the red block and the blue one is 0.2. If the blue block (10kg)rests on a frictionless surface, what will be its acceleration? F friction =  k n=  k mg=0.2*9.8=1.96 N (to the left) F red-blue =-F blue-red so force on blue block=1.96 N (to the right) F=ma so 1.96=10a a=0.196 m/s 2

9. PHY 231 9 example 6 A rocket is fired from a launching pad. The velocity AND acceleration of the rocket increase with time even though the thrust of the engine is constant. Why? a)The gravitational force becomes smaller because the mass of the rocket becomes smaller (losing fuel) b) The gravitational constant g becomes smaller if the rocket travels away from the earth c) While traveling away from earth, the gravitational pull from other planets/sun becomes stronger supporting the further acceleration d) Because the density of air becomes small, the friction becomes less

10. PHY 231 10 Revision: chapter 5  Work: W=Fcos(  )  x Energy transfer  Power: P=W/  tRate of energy transfer  Potential energy (PE) Energy associated with position.  Gravitational PE: mgh Energy associated with position in grav. field.  PE stored in a spring: 1/2kx 2 x is the compression of the spring k is the spring constant  Kinetic energy KE: 1/2mv 2 Energy associated with motion  Conservative force: Work done does not depend on path  Non-conservative force: Work done does depend on path  Mechanical energy ME: ME=KE+PE  Conserved if only conservative forces are present KE i +PE i =KE f +PE f  Not conserved in the presence of non-conservative forces (KE i +PE i )-(KE f +PE f )=W nc

11. PHY 231 11 example 7 k=100 N/m m=1 kg 1 cm A pendulum is pushed with initial velocity 0.1 m/s from a height of 1 cm. How far does it compress the spring? (assume m does not rise significantly after hitting the spring) Conservation of ME: (mgh+1/2mv 2 +1/2kx 2 ) initial = (mgh+1/2mv 2 +1/2kx 2 ) final 1*9.8*0.01+0.5*1*0.1 2 +0.=0.+0.+0.5*100*x 2 so x=0.045 m

12. PHY 231 12 example 8 A ‘smart’ student decides to save energy by connecting his exercise treadmill to his laptop battery. If it takes 70 J to move the belt on the treadmill by 1 meter and 50% of the generated energy is stored in the battery, how ‘far’ must the student run to use his 100 W laptop for free for 2 hours? Work done by student: W=70*d J Energy given to the battery 0.5W=35*d J 100 W laptop for 2 hours: 100 J/s*3600*2 s=7.2E+5 J 720 KJ 7.2E+5=35*d so d=(7.2E+5)/35=2.1E+4 m =21km!!!

13. PHY 231 13 example 9 45 0 A block of 1 kg is pushed up a 45 o slope with an initial velocity of 10 m/s. How high does the block go if: a) there is no friction b) if the coefficient of kinetic friction is 0.5. A) Conservation of ME: (mgh+1/2mv 2 ) initial = (mgh+1/2mv 2 ) final 0.+0.5*1*10 2 =1*9.8*h+0. So h=5.1 m B) Energy is lost to friction: (mgh+1/2mv 2 ) initial = (mgh+1/2mv 2 ) final +W friction [W=F  x=  n  x=  mgcos(45 o )h/sin(45 o )=0.5*1*9.8*h=4.9h] 0.+0.5*1*10 2 =1*9.8*h+0.+4.9h so h=3.4 m h

14. PHY 231 14 example 10 A crate of 50kg is starting to slide from a slope. When it reaches the bottom, it is caught by a spring with a spring constant of 1000 N/m. a) If the crate was originally at a height of 10 m and friction can be ignored, how much is the spring maximally compressed? b) if the frictional force is 100N and the length of the slope is 15m, what is the maximal compression? a)use conservation of mechanical E. at top: E tot =1/2mv 2 +mgh=mgh=50*9.8*10=4900 J when spring is maximally compressed: E tot = 1/2mv 2 +mgh+1/2kx 2 =1/2*1000*x 2 =500x 2 4900=500x 2 x=3.1 m b) W non-conservative =F friction *  x=100*15=1500N The mechanical energy just before the block hits the spring: 4900-1500=3400J 500x 2 =3500 x=2.6 m

15. PHY 231 15 example 11 A car (A) can accelerate from 0 to 10 m/s in 5 s. Another car (B) of the same mass as A can reach 8 m/s in 5 s seconds. What is the ratio of the power of car A to car B ( P A /P B )? A)10/8 B)8/10 C)1 D)10/5 E)8/5 F)100/64 P=W/  t since  t is the same for A and B, P~W. W=KE f -KE i =0.5m(v f 2 -v i 2 ) W A =0.5m10 2 W B =0.5m8 2 P A /P B =W A /W B =10 2 /8 2 =100/64

16. PHY 231 16 example 12 A 70-kg diver steps of a 10-m tower and drops, from rest straight down into the water. If he comes to rest 5.0 m below the water surface, determine the average resistive force exerted by the water. C A B (KE i +PE i )-(KE f +PE f )=W nc Choose the final rest point under water as the zero level. KE A =0 PE A =mgh=70*9.8*15=10290 J KE C =0 PE C =mgh=70*9.8*0=0 J W friction =10290 N=(Fcos  )  x=5F F=2058 N Maximum speed (just before hitting the water): KE A +PE A =KE B +PE B 0+10290=0.5*70*v 2 +70*9.8*5 v=14 m/s