v = v i + at a dv dt = d2 x dt 2 A sphere = 4πr 2 x = x i + v i t + 1 2 at2 x = r cos(θ) V sphere = 4 3 πr3 v 2 = v 2 i + 2a x F = ma R = v2 sin(2θ) g y = r sin(θ) r = x 2 + y 2 tan(θ) = y x a c = v2 r = rω2 F c = m v2 r = mrω2 v dx dt A circle = πr 2 F f = µf n Constant Symbol Approximate Value Gravitational Constant G 6.67 10 11 Nm 2 /kg 2 Acceleration of Gravity (near Earth) g 9.80 m/s 2 Speed of light in vacuum c 3.00 10 8 m/s Mass of Earth M E 5.97 10 24 kg Radius of Earth s Orbit R E 1.496 10 11 m Mass of Moon M M 7.35 10 22 kg Radius of Moon s Orbit R M 3.84 10 8 m 1
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Physics 204A Exam 2 October 10, 2016 Short-answer problems: Do any seven problems. Clearly indicate the problem you wish to skip. Six points each. 1. In lab, you measure force on a cart and the resulting position of the cart, and plot them on a graph as shown below. What is your best estimate of the mass of this cart? 1.2 1.0 0.8 Force (N) 0.6 0.4 0.2 0.0 0.0 0.5 1.0 1.5 2.0 Acceleration (m/s 2 ) 2. A heavy box is in the back of a truck, which is moving forward. The truck driver has just applied the brakes. Draw separate free-body diagrams for both the crate and the truck, showing all forces acting on each. 3
3. Bubba (m = 80 kg) is in an elevator which is traveling downwards at v = 10 m/s. The elevator takes t = 3.0 s to stop when it reaches the bottom. What is Bubba s weight while the elevator is stopping? 4. A truck with a heavy load has a total mass of M = 7500 kg. It is climbing a 15 incline at a steady speed of 15 m/s when the load falls off. (I hate it when that happens... ) The truck immediately begins to accelerate at 1.5 m/s 2. What was the mass m of the load? 4
5. For the two blocks shown below, the coefficient of static friction between the upper and lower block is µ s = 0.6. The coefficient of kinetic friction between the lower block and the floor is µ k = 0.2. Force F causes the two blocks to move across the floor a total distance x = 5 m, starting from rest. What is the least amount of time in which this motion can be completed without the top block sliding on the bottom block? Hint: you don t need to know F. 4 kg F 3 kg 6. A mass is supported by the system of pulleys shown. Assume that the pulleys are frictionless and (compared to m) massless. Find the tensions in each cord segment (T 1 through T 5 ) and the force F necessary to lift m at a constant speed. T5 T2 T3 T4 T1 m F 5
7. Mass m 2 is rotating in a circular path, radius r, on a frictionless horizontal table. It is attached to mass m 1 via a string of negligible mass, which is hanging below the table. At what angular velocity ω must m 2 rotate so that m 1 does not accelerate? r m2 m1 8. One satellite inadvertently put into orbit in the early 1960 s was a Hasselblad camera (m = 1.2 kg) accidentally dropped by one of the astronauts during a Project Gemini spacewalk. If the radius of the Hasselblad s orbit was 6500 km, and it orbited the Earth once every 90 minutes, what was the gravitational force on this camera? Compare this force with the weight of the camera on the Earth s surface. 6
Physics 204A Key Exam 2 October 10, 2016 Short-answer problems: Do any seven problems. Clearly indicate the problem you wish to skip. Six points each. 1. Answer: From Newton s 2 nd law, F = ma. So the slope of a Force-Acceleration graph will be the mass m. The slope of the graph shown is roughly 1/1.75, so m = 0.57. (Answers will vary, but should be close to that value.) FNtb FNgt Ffbt Box Ffgt Truck Fftb Fgb FNbt Fgt 2. Answer: 3. Answer: During the stop, Bubba s acceleration is a = v/ t = 10/3 (upwards). The forces on Bubba are gravity (downwards) and the normal force of the elevator s floor (upwards.) That normal force is his weight. The net force (up is positive) is then f net = F N F g ma = F N mg ( F N = ma + mg = weight = m g + (10/3) m/s 2) = 1050 N 4. Answer: The truck was moving at a steady speed, so the net force was zero. This means that the downhill component of gravity was equal to the force of the truck (technically friction), F t Mg sin θ = 0 Once the load m falls off, the truck accelerates because the new mass (M m) is less but the engine force F t is still the same. F t (M m)g sin θ = Ma But we know F t : Mg sin θ (M m)g sin θ = Ma mg sin θ = Ma = m = M a g sin θ m = 4440 kg 1
FNbt Ffbt F FNtb Fgt Fftb Fff FNf Fgb 5. Answer: Start with a free-body diagram. Here s the thing, though: we don t know F, and we don t need to know F! We just need the maximum acceleration of the bottom block, and once we have that we can use kinematics equations. The only horizontal forces on the bottom block are F ftb (the friction of the top block on the bottom block) and F ff (the friction between bottom block and floor.) a = F m = F ftb F ff m b a = µ sm t g µ k (m t + m b )g m b = 3.3 m/s 2 That s the maximum acceleration of the bottom block, so use kinematics equations with this maximum acceleration. x f = x i + v i t + 1 2 at2 6. Answer: x f = 1 2xf 2 at2 = t = a T 1 = mg = 1.75 s T 2 = T 3 = T 4 = 1 2 mg T 5 = 3 2 mg F = T 4 = 1 2 mg 7. Answer: The centripetal force m 2 rω 2 is provided by the string tension F T. Since m 1 is not accelerating, this tension must be equal to the weight of the hanging mass. F T = m 1 g = m 2 rω 2 ω = m1 g m 2 r 2
8. Answer: The angular velocity of the camera orbiting the Earth is 2π rad 5400 s The centripetal force is most easily given by It s weight on Earth is only slightly more: = 0.00116 rad/s F c = mrω 2 = 10.6 N mg = 11.8 N 3