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Download PHYSICS 231 Review problems for midterm 1 1 PHY 231
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PHYSICS 231 Review problems for midterm 1 PHY 231 1 Velocity (m/s) v(t)=v(0)+at x(t)=x(0)+v(0)t+0.5at2 Cut problem up in 1s pieces After 1 s: v(1)=0+0x1=0 x(1)=0+0x1+0.5x0X12=0 1 1 1.5 3 3 After 2 s: v(2)=v(1)+at=0+3x1=3 x(2)=x(1)+v(1)t+0.5at2 =0+0x1+0.5x3x12=1.5 What is the displacement at t=4 s. By drawing: Derive v(t) diagram from a(t) diagram: red line x(t) is area under v(t) diagram: PHY 231 After 3 s: v(3)=v(2)+at=3+2x1=5 x(3)=x(2)+v(2)t+0.5at2 =1.5+3x1+0.5x2x12=5.5 After 4 s: v(4)=v(3)+at=5-2x1=3 x(4)=x(3)+v(3)t+0.5at2 =5.5+5x1+0.5x(-2)x12=9.5 20 Cross straight a) Cross fast v? 6.50 m/s 3.30 m/s b) 6.50 m/s 3.30 m/s 3.30 m/s To cross straight: use picture a) Velocity of water Velocity of boat needed to cancel motion of water Total velocity of the boat (I.e. available in still water) Velocity ‘left over’ for crossing river v2+3.302=6.502 so v=(6.502-3.302)=5.6 m/s Time to cross river: t=x/v=262.0/5.6=46.8 s (use v=x/t) To cross fast: use picture b) V=6.5 m/s so t=x/v=40.3 s (but lands downstream) PHY 231 21 + a) b) c) d) Acceleration in vertical direction is ALWAYS g=–9.81 m/s2; TRUE See a) FALSE It is positive going up, negative going down: FALSE It is zero at the start and end and positive everywhere else: TRUE PHY 231 22 vx(0)=29.0 m/s 2.19 m x(t)=x0+v0xt = 0+29t=29t vx(t)=v0x = 29 y(t)=y0+v0yt-0.5gt2 = 2.19+0t-0.5x9.8t2=2.19-4.9t2 vy(t)=v0y-gt = 0-9.8t=-9.8t When ball hits ground: y(t)=0 so: 2.19-4.9t2=0 t=0.67s Use in x(t) equation: x(0.67)=29 x 0.67 = 19.4 m PHY 231 23 Acceleration in horizontal direction: use F=ma You only need the horizontal component, so: Fx=F x cos()=11.4xcos(21)=10.6 N a=F/m=10.6/5.22=2.04 m/s2 Now use v=v(0)+at=0+2.04x4.10=8.36 m/s PHY 231 24 1) Draw forces 2) The block has constant velocity, so no acceleration, so no net force! n=FgL Fg//=Fgsin() Ffr=kn FgL=Fgcos() Fg In direction parallel to the slope: F-Fg//-Ffr=0 so F-Fgsin()-kn=0 and… F-mgsin()-Ffr=0 18.9-1.12x9.81xsin(36.5)- Ffr=0 Ffr=12.4 N PHY 231 25 1) Draw forces 2) Since the block is not moving, no acceleration, no net force mg T=15N Note that the tension in the lower block must be trying to pull The block down-the rope cannot support weight and so not Produce an upward force by itself… Tupper-mg-Tlower=0 so Tupper=4.3x9.81+15=42.2+15=57.2 N PHY 231 26 n=mg Ffr=n= mg M T Fg=mg a) No acceleration: no net force. Ffr=T maximal Ffr=sn so T sn : TRUE b) k < s and acceleration will start if T> sn so FALSE c) At rest, so no net force: F=0: false. PHY 231 27 See LON-CAPA The force pulling M2 to the right is M2gsin=mgsin The force pullin M1 down is M1g=mg mg>mgsin so M1 wins and M2 goes left, and M1 goes down (both accelerate). Also T1=T2 a) Net force on M1: T-mg. M1 goes down, so gravity wins and M1g>T answer: B b) Must be equal: connects : answer C c) T2 wins from M2gsin (M2 goes left) so answer: B d) M2g=M1g and M1g>T so M2g>T answer B e) Equal (else rope breaks) PHY 231 28 Work is equal to the change in potential energy (assume That kinetic energy at start and stop is 0) W=mghf-mghi=471x9.81(24.7-0)=114127 J=1.14x105 J P=W/t=1.14x105/(2.50x60s)=761 J/s = 761 Watt PHY 231 29 335 km/hr= 93.1 m/s Work-energy theorem: MEi-MEf=(PE+KE)i-(PE+KE)f=Wnc PEi=mgh = 71.4 x 9.81 x 3490 = 2.44x106 J KEi=0.5mv2=0.5 x 71.4 x 93.12 = 3.09x105 J MEi=2.75x106 J PEf=mgh = 71.4 x 9.81 x 0 = 0 J KEf=0.5mv2=0.5 x 71.4 x 3.652 = 475 J MEf=475 J Wnc= 2.75x106 J – 475 J = 2.75x106 J PHY 231 30 L: length units T: time units Write out dimensions: L = bT3 + cT4 To get length units from the term bT3, b must have units L/T3 To get length units from the term cT3, c must have units L/T4 Answer: C. PHY 231 31 Acceleration is constant, so: x(t)=x(0)+v(0)t+0.5at2 = 0.5at2 Say the incline is d long. To cover the full length d: d=0.5at12 t1=(2d/a) To cover half the incline (d/2) d/2=0.5at22 t2=(d/a) t2/t1 = 1/2 = 0.71 so answer c. It takes more time to cover the first half (since the average velocity is lower) 32 PHY 231 The average speed = distance covered / time that it took Use: y(t)=y0+v0yt-0.5gt2 = 19.6t-0.5x9.8t2 vy(t)=v0y-gt = 19.6-9.8t 1) Total distance=twice the distance up At highest point: vy=0 so 19.6=9.8t t=2 y(2)=19.58 m distance covered: 2x19.58=39.16 2) Total time: when the ball returns: y(t)=0 19.6t-0.5x9.8t2=0 t=0 (start) or t=4 So average speed: 39.2/4=9.8 m/s Note: you could also have used just the way up…. PHY 231 33 Velocity: m/s a) x2/t units: m2/s not equal to m/s b) ½ x gt2 units: m/s2 xs2 = m not equal to m/s c) x/(t-x) units: m/(s-m): this is not possible and certainly not equal to m/s d) x/t units: m/s equal to velocity e) x/t2 units: m/s2 not equal to velocity PHY 231 34 225o 1350 Decompose in ‘horizontal’ (west-east direction) and ‘vertical’ (northsouth) components Horizontal Vertical Day 1 15*cos(225)=-10.6 15*sin(225)=-10.6 Day 2 10*cos(135)=-7.07 10*sin(135)=7.07 Sum -17.7 -3.53 Total displacement: PHY 231 35 PHY 231 36 PHY 231 37 Initially, the velocity is pointing up, but is decreasing in magnitude (speed is decreasing) since the gravitational force is slowing it down. This goes on until it reaches the highest point, where the velocity/speed equals zero. The ball than moves down: the velocity becomes negative, but the speed (not a vector, just a positive number) increases. So answer c is correct. PHY 231 38 x(t)=x0+v0t+0.5at2 v(t)=v0+at X(t)=20-vot-0.5gt2 with g=9.81 m/s2 At t=1.5 s, x=0, so 0=20-1.5vo-0.5*9.81*(1.52) Solve for vo =-5.98 m/s magnitude: +5.98 m/s answer b) PHY 231 39 10 km/h 5 km/h The boat will move under an angle determined by tan=5/10 -> =26.57o The tan=0.5 also is equal to (distance d)/(width of river=1 km) = d/1 So d=0.5 km. PHY 231 40 x(t)=x0+v0xt vx(t)=v0x = v0cos() y(t)=y0+v0yt-0.5gt2 v(t)=v0y-gt g=9.81 m/s2 v0y=v0sin() x(t)=x0+v0xt = 40cos(40o)t we don’t now t… y(t)=y0+v0yt-0.5gt2 = 0.2+40sin(400)t-0.5*9.8*t2 =0 at landing Solve this for t (quadratic equation): t=5.239 s Plug this into the equation for x(t): x(5.239)= 40cos(40o)5.239=161 m answer d) PHY 231 41 PHY 231 42 1 2 Object 1: Object 2: F=m1a, so F=m2a, so T-Ffr=m1a (moving to the right) Fg-T=m2a (note the – sign!!) m2g-T=m2a The frictional force Ffr=µkn=µkm1g (magnitude of normal force equals the gravitational force) So we have: Block 1 T-Ffr=m1a -> T-µkm1g=m1a -> T-0.1*40*9.8=40*a Block 2 m2g-T=m2a -> -T+10*9.8=10*a Sum to eliminate T 0-39.2+98=50a 43 So a=1.2 m/s2 The acceleration is the same for both masses The reading of the scale equals the normal force provided by the scale. Write down Newton’s law for the forces acting on you: F=ma n-mg=ma (normal force n is pointing up, gravitational force is pointing down) The elevator is accelerating upwards, so a>0 and thus: n>mg which means that the weight you read from the scale is larger than your nominal weight (i.e. when not accelerating), answer a) PHY 231 44 Newton’s law for motion parallel to the slope: F=ma -mgsin + Ffan =ma (down the slope is negative) -0.5*9.81*sin+2.45=0.5*a (cart is at rest, so a=0) -0.5*9.81*sin+2.45=0, solve for -> =300 answer a) PHY 231 45 PHY 231 46 Since the skier eventually stops, all the kinetic energy he had at the bottom of the slope was ‘taken’ (dissipated) by friction. So if we determine the kinetic energy at the bottom of the slope, we know the answer to the problem. To find the KE at the bottom of the slope we use conservation of energy: KEi+PEi=KEf+PEf KEi : kinetic energy at top = 0 (starts from rest) PEi : potential energy at top = mgh = 70*9.8*200=137 kJ PEf : potential energy at bottom = mgh=0 (h=0) So, KEf = PEi=137 kJ this is also equal to the energy dissipated by friction after the skier comes to a full stop. 47 PHY 231