|
AP Physics: Thurs SEP 24: Target: Learning Objective: Impulse & Mechanical Advantage Journal:
5000m 100
A 100,000 kg train climbs a 5,000 m mountain track at a speed of 20m/s at an angle of 10˚ with respect to the horizontal. 1. How much altitude does it gain? a. 868 m b. 1230 m c. 3410 m d. 4920 m 2. How much work was done?
3. How much power was used?
4. What is the mechanical advantage (MA) of the mountain?
1. Sin10˚(5000m) = 868 m 2. W = (f)(x) = DPE W = Dmgh = (100,000kg)(10N/kg)(868m) = 8.68 x 108 J 3. P = W/t P = 8.68 x 108 J / (5000m/20m/s) P = 3.47 x 106 W 4. MA = ratio of distances = 5000m/868m = 5.76
Chapter 4:
(forces)
*Free-body diagrams & Forces:
1. The three blocks are accelerating at 5m/s2 Draw Free Body diagrams & find the tensions in the lines.
T1 , T2 , & T3
![]() Fnet = ma à F = (60kg)(5N/kg)à 300N so, T1= 300N
Fnet on 30kg block = T1 - T2 = ma à Fnet = 300N - T2 = (30kg)(5N/kg) à 300N - 150N = T2 so, T2= 150N
Fnet on 10kg block = T2 - T3 = ma à Fnet = 150N - T3 = (10kg)(5N/kg) à 150N - 50N = T2 so, T2= 100N
2. What is the a. Mechanical Advantage b. Effort Force (Resistance force is 100N) c. Effort Distance (Resistance dist. is 2.0m) I. incline plane of 300 II. Wheel (2.4m) & Axel (0.2m) III. Pulley of 1 weight bearing line III. Pulley of 3 weight bearing lines IV. Pulley of 4 weight bearing lines V. 12m Lever: 2m on resistance side of fulcrum and 10m on effort side VI. 16m incline plane that is 4m high
I. a. Sin 300 = .5, ratio of distances so MA = 2 b. 50N (½ the force) c. 4.0 m (twice the resistance distance)
II. a. 2.4/0.2 = 12, so MA = 12 b. 8.33N (100/12) c. 24m (2m)(12)
III. a. MA = 1 b. 100N c. 2m
IV. a. MA = 4 b. 25N (100/4) c. 8m (2m)(4)
V. a. MA = 5 b. 20N (100/5) c. 10m (2m)(5)
VI. a. MA = 4 b. 25N (100/4) c. 8m (2m)(4)
3. A 170kg hot air balloon hovers at a constant altitude above the ground. The gondola while a 80kg person is hanging from a rope below the gondola. The person on the rope starts to ascend at constant rate at 1.0m/s. a. What is the direction of the gondola as he ascends? b. Speed of the balloon as he ascends? b. If the person climbs for 10 seconds and then stops, what is the change in the balloons altitude?
a & b.) Momentum (Mo) is conserved Mo up = Mo down mVup = mVdown à (80)(1) = (170)(Vdown) Vdown = - 0.471 m/s b. Person below went up 10m: PE is conserved (no KE) mghperson = mghgondola (80)(10)(10) = (170)(10)(y) y = 4.71m down
2nd way also: Distance = Vt = (0.471m/s)(10s) = 4.71m down
4. A trunk is pulled across a floor at a constant speed by a 200N force applied by a rope at an angle of 600 above the horizontal. If the trunk weighs 150N. The coefficient of friction is:
FK = Fhorz = (m) FN (m) FN = (Cos 60)(200)
(m) (150 - .87(150) = (.5)(200) (m) (150 - 130) = (.5)(200) m = 100 / 20 = 5
5. A man pushes a 10kg cart up a 300 incline at a constant speed. He applies 60N force parallel to the slope. a. What is the Normal force? b. What is the force down the plane? c. What is the force of Kinetic friction? d. What is kinetic m?
FN = (Cos q)(mg)
Fx = (Sin q)(mg)
a. FN = (Cos 30)(mg) = (.87)(10)(10) = 87N b. Fx = (Sin 30)(mg) = (.5)(10)(10) = 50N (no friction) c. Constant speed means no acceleration (Balanced forces) Applied force = Fx + Friction 60N = 50N + 10N (of Kinetic friction) d. FK = mFN = m (Cos 30)(mg) FK = 10N = m (87N) m = .115
6. The block is resting on an incline plane (static m = 0.30 and kinetic m = 0.18). The angle is increased until the block starts to move. a. What is the angle (block starts to move)? b. What is the block’s acceleration at this angle? c. What is the speed of the block when it hits the ground (the length of the incline plane = 4.0m)? d. After reaching the floor, the block slides with the same kinetic m (0.18). How far does it slide?
Fx = (Sin q)(mg): FS = mSFN: FN = (Cos q)(mg) a) Fx = FS à (Sin q)(mg) = mS(Cos q)(mg) à (Sin q) = mS(Cos q) à mS = (Sin q)/(Cos q) = Tanq mS = Tanq .3 = Tanq à Tanq -1 (.3) = 16.70
b. Fx = (Sin q)(mg) , FK = mKFN , FN = (Cos q)(mg) Fnet = ma Fx – FK = ma (Sin q)(mg) – mK(Cos q)(mg) = ma à (Sin 16.7)(g) – 0.18(Cos 16.7)(g) = a à (.287)(10) – 0.18(.958)(10) = a à (2.87) – (1.72) = a à a = 1.15m/s2
Vf2 – 0 = 2(1.15)(4) Vf = √ 9.2 = 3.03m/s
mK(mg) = ma à mK(g) = a
Vf2 – Vo2 = 2aDs 0 – (3.03)2 = 2 (-1.8)s à s = 5.6m
5kg
12. A 5kg block hangs over a table and pulls a 4kg block on the table with mK = 0.4. What is the: a. Net force? b. acceleration? c. tension in line?
Fnet = m5g – (mm4g) Fnet = 50 - (.4)(40) à 16N Fnet = 50 - 16N = 34N
34 = 9a à a = 3.8m/s2
W – T = m5a 50 –T = (5)(3.8) T = 50 - 18.9 à 31.1N
13. A 10.0m board (100kg) is hanging 2.5m over the edge of a platform (that is above the water). Jimmy wants to take a good look at the water. If Jimmy (120kg) starts walking towards the end of the board, will he make it all the way without falling in? If not, how far can he walk on the board before both he and the board fall into the water below?
GOC of board is 5m à 2.5m from edge Torque from platform edge: fd (ccw) = fd (cw) (1000)(2.5) = 1200(x) x = 2.08m Yes, sorry to say, Jimmy would fall!
|
||||