PHY PHYSICS 231 Lecture 39: Review Remco Zegers Question hours: Monday 15:00-17:00 Cyclotron building seminar room

PHY chapter 4 A person is pushing an ice-sledge of 50kg over a frozen lake with a force of 100N to the east. A strong wind is pushing from the south-west and produces a force of20N on the sledge. a) What is the acceleration of the sledge? b) if the coefficient of kinetic friction is 0.05, what is the acceleration? a)force Horizontal vertical person wind 20cos(45) 20sin(45) sum = Total force:  ( )=115 N F=ma 115=50a a=2.3 m/s 2 b)F friction =  n=0.05*m*g=0.05*50*9.8=24.5N  F=ma =50a a=1.8 m/s 2 make sure you understand

PHY chapter 5 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 =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: =3400J 500x 2 =3500 x=2.6 m

PHY chapter 6 Two objects collide head on. Object 1 (m=5kg) has an initial velocity of 10m/s and object 2 (m=10 kg) has an initial velocity of -8 m/s. What is the resulting velocity a)of the combined object if the collision is fully inelastic? b)of each of the objects if the collision if fully elastic? a)inelastic: only conservation of momentum. m 1 v 1i +m 2 v 2i = (m 1 +m 2 )v f 5*10+10*(-8)=15*v f v f =-2m/s b) elastic: conservation of momentum and kinetic energy momentum: m 1 v 1i +m 2 v 2i = m 1 v 1f +m 2 v 2f 5*10-10*8=5v 1f +10v 2f kinetic energy: (v 1i -v 2i )=(v 2f -v 1f ) 10-(-8)=v 2f -v 1f -30=5v 1f +10v 2f & 18=-v 1f +v 2f v 2f =18+v 1f -30=5v 1f +10(18+v 1f ) v 1f =-14 v 2f =4 m/s

PHY chapter 10 An ideal gas in a volume of 5m 3 at 1 atm is compressed to a volume of 2m 3 and the pressure changes to 1.5 atm. If the original temperature was 27 o C, what is the change in internal energy of the gas? original: pV/T=nR 1.0x10 5 *5/(273+27)=1.67x10 3 =nR p 1 V 1 /T 1 = p 2 V 2 /T 2 T 2 =p 2 V 2 T 1 /p 1 V 1 =180K U=3/2nRT  U=3/2nR  T=3/2*1.67x10 3 *( )=-3x10 5 J faster: U=3/2nRT=3/2pV  U=3/2(p 1 V 1 -p 2 V 2 )=3/2(1.5x10 5 *2-1x10 5 *5)=-3x10 5

PHY Chapter 11 A cube of ice with mass of 1 kg and a base area of 0.01m 2 is placed on top of a aluminum plate of 1cm thickness. The plate is heated from below and the temperature just below the plate is 50 0 C. Given k Al =238 J/sm 0 C and L ice =3.33x10 5 J/kg, how long does it take to melt all the ice, assuming no heat is lost. Q=mL=1*3.33x10 5 =3.33x10 5 J P=kA  T/  x=238*0.01*50/0.01=1.19x10 4 J/s t=Q/P=28 s

PHY Chapter 11 A 0.2 kg aluminum plate, initially at 20 0 C slides down a 15-m long surface, inclined at 30 0 with the horizontal. The force of kinetic friction exactly balances the component of gravity down the plane. If 90% of the mechanical energy of the system is absorbed by the aluminum, what is the temperature increase at the bottom of the incline? (c Al =900 J/kg 0 C). F g =mg, along the slope: F g// =mgsin  =0.2*9.8*0.5=0.98N W froction =F  x=0.98*15=14.7 J 90% given to the plate: 0.9*14.7=13.2 J Q=cm  T  T=Q/cm=13.2/(900*0.2)  T= C.

PHY Chapter V(m 3 ) P(atm) Consider the process in the figure. a)How much work is done on the gas? b)What is the change in internal energy? c) how much heat was added to the system? a)Work: area under the graph: 1x x10 5 =1.5x10 5 J work done on the gas: -1.5x10 5 J b)  U=3/2  (nRT)=3/2  (pV)=3/2(p 2 V 2 -p 1 V 1 )= =3/2(1x10 5 *2-1x10 5 *1)=1.5x10 5 J c)  U=Q+W 1.5x10 5 =Q-1.5x10 5 Q=3x10 5 J

PHY Chapter 12 An engine is operated between a hot and a cold reservoir with Q hot =400J and Q cold =300J. a) what is the efficiency of the engine? The engine is modified and becomes a carnot engine. As a result the efficiency is doubled. b) what is the ratio T cold /T hot. c) what is the maximum efficiency of this engine? a)efficiency=1-Q cold /Q hot =1-300/400=0.25 b)new efficiency: 0.5=1-T cold /T hot T cold /T hot =0.5 c)0.5 (Carnot engine has maximum efficiency).

PHY Chapter 12 A block of ice of 1 kg at 0 0 C is melted (L=3.33x10 3 J/kg). What is the change in entropy?  S=Q/T=Lm/T=3.33x10 3 *1/273=12.2 J/K

PHY Chapter 13 I attach a 2.0 kg block to a spring that obeys Hook’s law and supply 16J of energy to stretch the spring. After releasing the block, it oscillates with a period of 0.30 s. What is the amplitude of the oscillation? Energy stored in the spring: 1/2kx 2 =16J When oscillating, the velocity is zero when the mass is at maximum amplitude, so 1/2kx 2 +1/2mv 2 =1/2kA 2 =16J A=  (2*16/k)  =  (k/m)  =2  f=2  /T so T=2  (m/k) 0.3=2  (2/k) so k=877 N/m and thus A=  (32/877)=0.19m

PHY Chapter 13 Transverse waves travel with a speed of 200 m/s along a copper wire that has a diametr of 1.5mm. What is the tension in the wire? (  copper =8.93 g/cm 3 ). v=  (F/  ) so F=v 2  v=200m/s  =mass of wire per unit length (1 meter) =density*volume: 8.93x10 3 kg/m 3 *(  r 2 *1)  =8.93x10 3 *1.77x10 -6 =1.58x10 -2 kg/m (diameter-> radius) F=632N

PHY Chapter 14 A football fan upset that his team is losing tosses out his battery powered radio out of a 20m high window. What is the frequency of the sound from the radio just before it hits the ground, relative to the frequency when it just drops out the window (assume: initial velocity 0). f’=f(v+v 0 )/(v-v s ) v: velocity of sound (343 m/s) v 0 : velocity observer (0 m/s) v s : velocity source. v s (t)=v 0 +at=0-9.8t x(t)=20-0.5gt 2 0=20-0.5*9.8*t 2 t 2 =40/9.8 t=2s so v=-19.6 s radio is moving away from the observer, so negative sign. f’/f=v/(v-v s )=343/(343-(-19.6))=0.95

PHY Chapter 14 If the distance between a point sound source and a dB detector is increased by a factor of 4, what will be the reduction in intensity level? Intensity~1/(distance from source) 2 I~1/r 2 I 2 /I 1 =r 1 2 /r 2 2 =1/4 2 =1/16  =10log(I/I 0 ) I 0 =1x W/m 2  2 -  1 =10[log(I 2 /I 0 )-log(I 1 /I 0 )]=10log(I 2 /I 1 )=10log(1/16)  2 -  1 =-12dB

PHY Chapter 8 An 800-N billboard worker stands on a 4.0-m scaffold supported by vertical ropes at each end. If the scaffold weighs 500N and the worker stands 1.0m from one end, what is the tension in the ropes? 1.0m Net force in vertical direction must be 0: F L +F R -w man -w scaffold =0 F L +F R =F L +F R -1300=0 Net torque must be zero. Choose 0 at ‘R’. F L *4-800*1-500*2=4F L -1800=0 F L =1800/4=450N so F R =1300-F L =850N. LR

PHY Chapter 7 A roller coaster, loaded with passengers, has a mass of mass of 2000 kg. The radius of the curvature of the track at the bottom point of the dip is 24m. If the vehicle has a speed of 18 m/s at this point, what force is exerted on the vehicle by the track? two forces: gravitational force and centripetal force. gravitational force: mg=2000*9.8=19600N centripetal force: mv 2 /r=2000*18 2 /24=27000N Summed: 46600N.