Physics 18 Spring 2011 Homework 3 - Solutions Wednesday February 2, 2011

Transcription

1 Phsics 18 Spring 2011 Hoework 3 - s Wednesda Februar 2, 2011 Make sure our nae is on our hoework, and please bo our final answer. Because we will be giving partial credit, be sure to attept all the probles, even if ou don t finish the. The hoework is due at the beginning of class on Wednesda, Februar 9th. Because the solutions will be posted iediatel after class, no late hoeworks can be accepted! You are welcoe to ask questions during the discussion session or during office hours. 1. Astronauts in apparent weightlessness during their sta on the International Space Station ust carefull onitor their asses because significant loss of bod ass is known to cause serious edical probles. Give an eaple of how ou ight design equipent to easure the ass of an astronaut on the orbiting space station. There are all sorts of was that ou ight easure the ass of an astronaut. One wa would be to put the astronaut on a spring with known spring constant, k, set hi oscillating, and easure the oscillation frequenc, which depends on the ass. Another wa ight be to attach hi to that spring, and spin hi in a circle. Then ou can easure the epansion of the spring, which is also equal to the centripetal force. In this case, ou can relate the astronaut s ass to the epansion of the spring. 1

2 2. A net force of (6.0 N) î (3.0 N) ĵ acts on a 1.5 kg object. Find the acceleration a. What is the agnitude of the acceleration, a? Newton tells us that F = a, where F is the applied force, is the ass of the object, and a is the acceleration the ass eperiences. So, the acceleration is just a = F /. So, we just find a = F = 6.0î 3.0ĵ 1.5 = ( ) 4.0î 2.0ĵ /s 2. This is our acceleration. The agnitude is just a = a 2 + a 2 = = 20 = 4.47 /s 2. 2

3 3. Seat belts and air bags save lives b reducing the forces eerted on the driver and passengers in an autoobile collision. Cars are designed with a cruple zone in the front of the car. In the event of an ipact, the passenger copartent decelerates over a distance of about 1 as the front of the car cruples. An occupant restrained b seat belts and air bags decelerates with the car. B contrast, an unrestrained occupant keeps oving forward with no loss of speed (Newton s first law!) until hitting the dashboard or windshield. These are unielding surfaces, and the unfortunate occupant then decelerates over a distance of onl about 5. (a) A 60 kg person is in a head-on collusion. The car s speed at ipact is 15 /s. Estiate the net force on the person if he or she is wearing a seat belt and if the air bag deplos. (b) Estiate the net force that ultiatel stops the person if he or she is not restrained b a seat belt or air bag. (c) How do these two forces copare to the person s weight? For a constant acceleration we can relate the initial and final velocities as v 2 f = v2 i + 2a. Since the final velocit is zero, then the acceleration is a = v2 i. Thus, the 2 force acting on the passenger is F = a = v2 i. We just need to plug in the different 2 distances for each case. (a) When the air bags deplo and the seat belts hold, then the stopping distance is = 1, and so F = v2 i 2 = = 6750 N, where the negative just eans that it s pushing the passenger back into the seat. (b) If there are no restraints, then = 5, or = 0.005, and so F = v2 i 2 This is uch bigger than the restrained case! = = N! (c) The passenger weighs W = g = = 588 N, and so the result fro part (a) is about 11.5 ties bigger than their weight, while the result fro part (b) is about 2300 ties their weight! It s clear which is a ore survivable ipact! 3

4 4. A 35.0-kg traffic light is supported b two wires as in the figure. (a) Draw the light s free-bod diagra and use it to answer the following question qualitativel: Is the tension in wire 2 greater than or less than the tension in wire 1? (b) Verif our answer b appling Newton s laws and solving for the two tensions. (a) The free-bod diagra is seen in the figure to the right. You ll notice that the vectors do add to zero, which, as we know, is the case if the light isn t accelerating. That is, T 1 + T 2 + = 0. Fro the diagra, it sees prett clear that the tension T 2 > T 1. (b) Now, let s check the ath. The su of the forces in each direction has to be zero. Since gravit points down, it doesn t contribute to the forces along the direction. So, F = T 1 cos 30 T 2 cos 60 = T 2 T 1 30 The two forces cancel out in the horizontal direction. So, we can epress T 2 in ters of T 1, which gives T 2 = cos 30 cos 60 T 1 = 3/2 1/2 T 1 = 3T 1, and so, indeed, T 2 > T 1, as we guessed. 4

5 5. A 65-kg student weights hiself b standing on a force scale ounted on a skateboard that is rolling down an incline, as shown in the figure. Assue there is no friction so that the force eerted b the incline on the skateboard is noral to the incline. What is the reading on the scale if θ = 30? The reading that the scale gives is equal to the noral force, the force that the scale eerts on the skater. We can draw F n a free-bod diagra, choosing our coordinates such that the ais points along the rap, as in the figure to the right. Then, the noral force points entirel along the direction. The su of the forces in the direction has to be zero, since the skater is accelerating along the rap (i.e., in the direction). So, F = F n cos 30 = 0. Since the gravitational force is = g, then 30 F n = g cos 30 = 3 2 g = 3 (65)(9.8) = 552 N. 2 So, the scale reads 552 N. 5

6 6. A block of ass slides across a frictionless floor and then up a frictionless rap. The angle of the rap θ and the speed of the block before it starts up the rap is v 0. The block will slide up to soe height h above the floor before stopping. Show that h is independent of and θ b deriving an epression for h in ters of v 0 and g. Since the floor is frictionless, the block slides along the floor at a constant speed, v 0. Once it starts going up the rap, then it s slowed b gravit pulling it back down. We can draw a force diagra for the block, orienting our coordinate sste along the rap,. In this case, the noral force points entirel along. Suppose that the block oves along the rap a distance d, as in the figure. Then, fro the geoetr, it is up a height h = d sin θ. So, if we can figure out how far it goes along the rap, then we ll know our answer. We can figure out how far along the rap it goes using the kineatic equations. Since we don t have the tie it takes to travel up the rap, we can use the velocit-distance equation, v 2 f = v 2 i + 2a, where a is the acceleration along the direction (which isn t g!). When the block reaches its aiu height, it stops oving, and so v f = 0. So, if it started with an initial velocit v 0, and travels a distance = d, then solving for d gives d = v2 0 2a. What s a? We can deterine this fro Newton s laws. Since the block is accelerating along, and the noral force points entirel along, then F = g sin θ = a. Thus, a = g sin θ. So, d = v2 0. Plugging in our epression for h gives 2g sin θ h = d sin θ = v2 0 2g sin θ sin θ = v2 0 2g, which is independent of the ass and angle, as advertised. 6 d θ F n θ h

7 7. A block of ass is being lifted verticall b a unifor rope of ass M and length L. The rope is being pulled upward b a force applied to its top end, and the rope and block are accelerating upward with an acceleration of agnitude a. Show that the tension in the rope at a distance (where < L) above the block is given b (a + g) [ + (/L) M]. The sste is seen in the figure to the right. The total aount of ass below the point is the ass of the block,, plus the ass of the segent of rope of length. The trick is figuring out how uch that ass is. Since the ass is unifor, ever eter of the rope has the sae ass. So, M/L is a constant nuber, and sas how uch ass a unit length of the rope carries. So, if we take this nuber, and ultipl it b the length,, this this tells us how uch ass in in the segent. So, segent = M. Thus, the total ass below the point L is L M T (+(M/L)) M total = + M L. Now that we know what the ass is, we can deterine the tension fro a free-bod diagra, as seen above. The su of the forces along is F = T M total g = T ( + ML ) g. But, the force is the ass ties the acceleration, and the ass is the total ass of the block and segent, M total. So, setting F = ( + M L ) a and solving for T gives T = ( + ML ) (g + a). 7

8 8. The figure shows a 20-kg block sliding on a 10-kg block. All surfaces are frictionless and the pulle is assless and frictionless. Find the acceleration of each block and the tension in the string that connects the blocks. We can draw free-bod diagras for each block, as seen to the left. The diagra for the 20 kg block is first, while the 10 kg is on the right. The top block has three forces acting on it: the tension pulling it along, the gravitational force pulling it down, and the noral force pushing it up along. The botto block is a bit ore coplicated, having four forces. There is still the tension, the gravitational force, and the noral force, but the top block is sitting on the botto block, weighing it down a bit. So, there s an etra force, F 20. F n T F n T F 20 Let s start with the top block. Since the acceleration is in the direction, F = 20 a. So, working out the forces in the direction gives F = T 20 g sin 20 = 20 a, where we let the acceleration be negative, since the top block will likel slide backwards, since it s heavier. Working out the forces for the second block gives F = T 10 g sin 20 = + 10 a, where the acceleration is positive because it s opposite to that of the top block (if the top block slides down, then the botto one is sliding up, etc.). So, solving the top for the 8

9 tension and substituting in to the second gives ( ) g sin 20 = ( ) a. Solving for a gives ( ) a = g sin This is the acceleration. We can find the tension b plugging this back into our force equation for, sa, the botto block. ( ) T = 10 a + 10 g sin = 10 g sin g sin 20, or, So, plugging in the nubers gives ( a = T = g sin 20. ) g sin 20 = ( ) (9.8) sin 20 = 1.12 /s So, the botto block slides up at 1.12 /s 2. The top block slides down at the sae acceleration. Furtherore, T = 2 ( ) g sin 20 = 2 (9.8) sin 20 = 45 N

10 9. A 2.0 kg block rests on a frictionless wedge that has a 60 incline and an acceleration a to the right such that the ass reains stationar to the wedge. (a) Draw the free-bod diagra of the block and use it to deterine the agnitude of the acceleration. (b) What would happen if the wedge were given an acceleration larger than this value? Saller than this value? (a) The rap is accelerating to the right. This pushes against the block leading to a noral force. So, there are two forces acting on the block, the noral force and the gravitational force. Since the acceleration is in the direction, let s work out the su of the forces in the direction, F = a. F n F = F n cos 30 = a. So, in order to deterine the acceleration, we need the noral force, which we can deterine fro the forces along the direction. Since there is no acceleration in the direction, F = 0. Now, F = F n sin 30 g = 0 F n = g sin 30. Plugging this in for our acceleration gives 30 a = F n cos 30 = g cot 30 = (9.8) cot 30 = 17 /s 2. (b) The noral force coes fro the acceleration. An acceleration greater than 17 /s 2 would provide a bigger noral force than needed to balance the gravitational force. If this is the case, then, because there is a vertical coponent to the noral force, which is bigger than the gravitational force, there is an acceleration upwards, due to the forces not balancing. So, the block would slide up the rap. Conversel, if the acceleration was less than 17 /s 2, then the noral force wouldn t quite balance the gravitational force, and the block would slide down the rap. 10

11 10. Elvis Presle has supposedl been sighted nuerous ties since his death on August 16, The following is a chart of what Elvis s weight would be if here were sighted on the surfaces of other objects in our solar sste. Use the chart to deterine: (a) Elvis s ass on Earth, (b) Elvis s ass on Pluto, and (c) the free-fall acceleration on Mars. (d) Copare the free-fall acceleration on Pluto to the free-fall acceleration on the oon. Planet Elvis s Weight (N) Mercur 431 Venus 1031 Earth 1133 Mars 431 Jupiter 2880 Saturn 1222 Pluto 58 Moon 191 (a) The weight, W, of an object on a planet with gravitational acceleration g is just W = g. So, on Earth, Elvis has a ass = W/g = 1133/9.8 = 116 kg. (b) Because the ass is an intrinsic propert of an object, it doesn t change fro place to place. So, Elvis s ass on Pluto is eactl the sae as his ass on Earth, = 116 kg. (c) The free-fall acceleration on Mars, g Mars can be found fro Elvis s weight on Mars, divided b his ass, g Mars = W Mars / = 431/116 = 3.72 /s 2. (d) The free-fall acceleration on Pluto, g Pluto, is again, Elvis s weight on Pluto, divided b his ass, g Pluto = W Pluto /. The free-fall acceleration on the Moon is, b the sae reasoning, g Moon = W Moon /. Solving the second for the ass, and plugging it into the first gives ( ) WPluto g Pluto = g Moon. W Moon Plugging in the weights tells us that g Pluto = (58/191)g Moon = 0.3g Moon. So, the free-fall acceleration on Pluto is less than 1/3 that of the Moon! 11