Physics 140 Assignment 3 (Mechanics) his assignment must be handed in by 1 noon on hursday 0 th April 000. You can hand it in at the beginning of the lecture on that day or you may hand it to your laboratory demonstrator beforehand if you wish. You should write your answers in the space provided on these sheets. Space is also provided for you to show steps in your calculations. You should indicate any physical laws that you have used and any assumptions that you have made. Student number Your name Day of the week for your lab class Name of your laboratory demonstrator SOLUIONS 1. Five fat dictionaries, each 7.5 cm thick and each.5 kg, are resting side by side flat on a table 1.0 m high. How much work would it take to stack them one atop the other? Work done to raise a dictionary by height h W h Let the thickness of one dictionary be d hen total work done sum of work done on each dictionary (d + d + 3d + 4d) 10d 10.5 9.80 0.075 J 18.375 J Work required would be 18 J. A fairly small asteroid (950 kg) out in deep space is to be accelerated from rest up to 15 m/s. How much work will be required to do this? Use the Work - Energy heorem Net work done on object Change in kinetic energy of object (Net work done work done by net force) Work required final kinetic energy - initial kinetic energy 1 1 1 0 950 15 106875 J 1 Work required will be 11 10 4 J
3. While running, a person dissipates about 0.6 J of mechanical energy per step per kilogram of body mass. If a 60-kg person develops a power of 70 W during a race, how fast is the person running? Assume that a running step is 1.5 m long. Power Energy transformed time While running, person is transforming biochemical energy into mechanical energy which in turn is completely dissipated as heat, sound etc (because person is not accelerating). Energy transformed in one step ime for one step Power developed by person Speed distance step length step length power developed time time for one step Energy transformed in one step 15. 70. 917 m/s 06. 60 Speed of person is 3 m/s or 10 km/h (1 sig fig) 4. A spring hanging from a support has a hook at the bottom end. When a 1.5 kg mass is hung on the spring by gently lowering it onto the hook, the hook moves to a new position 15 cm below its original position. What is the spring constant of the spring? Mass is stationary and not accelerating. Net force on mass is zero hus force of spring on mass weight of mass F s 15 cm F s Force of spring on mass -kx (minus sign indicates that force is in opposite direction to the displacement i.e. the force is upwards) kx 15. 980. So k 98 N/m x 015. Spring constant is 98 N/m 5. A 45.0-kg girl is standing on a 155-kg plank. he plank, originally at rest and lying along a northsouth axis, is free to slide on a frozen lake, which is a flat, frictionless surface. he girl begins to walk northwards along the plank at a constant velocity of 1.50 m/s relative to the plank. (a) What is her velocity relative to the ice surface? (b) What is the velocity of the plank relative to the ice surface? Before Girl-plank system can be considered as isolated since ice surface is frictionless. here are no external forces with horizontal components. By law of conservation of linear momentum, total momentum of girl-plank system remains constant. After i.e. m g v g + m p v p 0 where v g is velocity of girl and v p is vel of plank in reference frame of ice surface. If v g is vel of girl in reference frame of moving plank then v g v g + v p (vector addition)
NB: in one dimension only need to consider magnitudes and signs of vectors. Solving the above two simultaneous equations gives g g+ mp( vg vg ) 0 p g 155 1. 50 vg 1. 163 m/s + mp 45. 0 + 155 155 1. 50 and vp vg vg 1. 50 0. 3375 m/s 45. 0 + 155 (a) Velocity of girl relative to ice surface is 1.16 m/s northwards (b) Velocity of plank relative to ice surface is 0.338 m/s southwards 6. Water is to be pumped to the top of the Physics Building which is 31 m high. What gauge pressure is needed in the water line at the base of the building to raise the water to this height? Pressure at base pressure due to weight of column of fluid + atmospheric pressure Gauge pressure at base Pressure at base - atmospheric pressure ρgh 1000 9.80 31 Pa 303800 Pa Gauge pressure needed is 3.0 10 5 Pa 7. A ferry boat is 4.10 m wide and 6.0 m long. When a large loaded truck pulls onto it, it sinks 4.0 cm in the water. What is the mass of the truck? d F B ferry air water For ferry width, w, length l, and mass M F that floats with a draught d, the buoyant force F B is given by F wldρg M g B F When truck mass M drives onto ferry the ferry sinks by a distance s. hus new buoyant force F wl( d + s) ρg ( M + M ) g B F ( ) F F M + M g M g M g wlρgs B B F F M wlρs 4. 10 6. 0 1000 0. 040 kg 1016.8 kg Mass of truck is 1.0 10 3 kg
8. An 0.0080-kg bullet is fired into a 0.095-kg block of wood at rest on a horizontal surface. After impact, the block slides 8.5 m before coming to rest. If the coefficient of friction between the block and the surface is 0.6, find the speed of the bullet before impact. After collision m M F fr F N d Before collision Before collision otal momentum before collision total momentum after collision i.e. ( m + v (m+g After collision F µ F µ ( m+ g fr N Net force on block F µ ( m+ g - µ ( m + 0 1 ( m + v v µ fr Net work done on bloch & bullet µ ( m+ Net work done KE (work - energy theorem) + m M v m µ + 0. 008 0. 095 0. 008 0. 6 9. 80 8. 5 18. 7 m/s Muzzle speed of bullet was 1 10 m/s (1 sig fig) 9. Water at a pressure of 3.0 x 10 5 Pa flows through a horizontal pipe at a speed of 1.0 m/s. If the pipe narrows to one-fourth its original diameter, find (a) the flow speed in the narrow section and (b) the pressure in the narrow section. A 1 A v 1 v Water is almost incompressible va va 1 1 v v A 1 πr 1 v1 A π 4 ( r ) Using Bernoulli's equation 16v 16 m/s 1 P1 + 1 ρv1 P + 1 ρv P P 1 1 ρv 1 1 ρv P 1 1 ρ( v1 16v1 ) 5 3. 0 10 + 0. 5 1000( 55 1. 0 ) 17500 Pa + + ( ) (a) Flow speed is 16 m/s (b) Pressure is 1.7 10 5 Pa NB: Strictly speaking these answers should be to 1 sig fig since diameter ratio given to 1 sig fig only
10.Calculate the angular momentum of the Earth that arises from its orbital motion around the Sun. he mass of the Earth is 6.0 x 10 4 kg and the distance of the Earth from the Sun is 1.5 x 10 11 m (NB you will need to read the chapter in Giancoli about rotational motion to answer this question). Moment of inertia of Earth about Sun, I mr Angular speed of Earth around Sun, π ω where 1 year Angular momentum of Earth arising from orbital motion around Sun, L, is given by mr π L Iω ( ) 6. 0 10 1. 5 10 π kg m s 60 60 4 365 69. 10 4 11 40 kg m s -1-1 Angular momentum of Earth due to its motion around Sun is.7 10 40 kg m s -1