Phyic 6A Practice Midter # olution or apu Learning Aitance Service at USB
. A locootive engine of a M i attached to 5 train car, each of a M. The engine produce a contant force that ove the train forward at acceleration a. If 3 of the car are reoved, what will be the acceleration of the horter train? We jut ue ewton nd law here: et force = (total a x (acceleration Initially, the total a i 6M (5 car, plu the engine After reoving 3 car the total a i 3M ( car, plu the engine In both cae the et force i the ae (the engine didn t change. Here i the forula in both cae: Initial = (6M(a inal = (3M(a final Setting thee equal give a final = a Another way to thin about thi one i that ince the a i cut in half, the acceleration ut double. or apu Learning Aitance Service at USB
. Two boxe are placed next to each other on a ooth flat urface. Box A ha a g and Box B ha a 3 g. A contant horizontal force of 8 i applied to Box A. ind the force exerted on Box B. We can thin of thi a a ingle box with total a 4g. Then uing = a we get the acceleration of the whole yte. 8 a 4 yte g a yte A B ow we do the ae thing, but jut for box B: g B 3 B 6 Reeber Box B ha a a of only 3g or apu Learning Aitance Service at USB
3. Two boxe are placed next to each other on a flat urface. Box A ha a g and Box B ha a 3g. The coefficient of friction are 0.3 and 0.4 for inetic and tatic friction, repectively. A contant horizontal force of 8 i applied. ind the acceleration of Box B. We can thin of thi a a ingle box with total a 4g. Then uing = a we get the acceleration of the whole yte. We need to account for friction, o firt find the axiu force of tatic friction: tatic,ax tatic,ax (0.4 (4 g 9.8 5.68 8 A B tatic Thi friction force i ore than the 8 force trying to ove the boxe, o they are held in place by friction. Thu the acceleration of both boxe i 0. *ote that the actual force of friction holding the boxe in place i only 8 jut enough to eep the fro oving. or apu Learning Aitance Service at USB
4. A crate filled with boo ret on a horizontal floor. The total weight of the crate and boo i 700. The coefficient of friction are 0.35 for inetic and 0.45 for tatic. A force of 450 i applied to the crate at an angle of 0 below the horizontal. ind the acceleration of the crate. We need to find the coponent of the 450 force: oral x y 450 450 co(0 in(0 43 54 friction x We need to find the noral force, o we can deterine whether tatic friction will hold the crate in place. or thi, we ue the fact that the force in the y-direction cancel out: net,y noral noral 854 700 54 0 Thi give a axiu force of tatic friction = (0.45(854 = 384 oparing thi with the x-coponent of the applied force, we ee that thi i not enough friction to hold the crate in place, o the actual friction force will be inetic. friction, = (0.35(854 = 99 ow we can ue =a again for the x-direction force: y 700 700 net,x x friction, a 43 99 ( a a. 74 9.8 or apu Learning Aitance Service at USB
5. Bloc A and B and are connected by a ale tring and placed a hown, with Bloc A on the upward lope of the 60 incline, Bloc B on the horizontal urface at the top and Bloc on the downward lope of the 60 incline. The tring pae over a frictionle, ale pulley. The incline are frictionle, but the horizontal urface ha coefficient of inetic friction 0.45. Bloc A and B have a 5g. The yte accelerate at /. ind the a of Bloc. B B T a= / T T B g T A A A gcoθ gcoθ ginθ A ginθ A g g Here i the force diagra for thi proble. The bloc will ove together, o we can conider the otion of the entire yte in one forula. Our axi yte will be choen to coincide with the tring. With thi choice, our forula i: gin A gin yte otice that the tenion force cancel out. We are auing the tring and pulley are ale, o the tenion i the ae throughout the tring. Thu the forward tenion T on A i canceled by the bacward tenion T on B, and iilarly the T cancel out. Or we ay that the tenion are internal force, and thu do not affect the acceleration of the entire yte. ( A B a or apu Learning Aitance Service at USB
5. Bloc A and B and are connected by a ale tring and placed a hown, with Bloc A on the upward lope of the 60 incline, Bloc B on the horizontal urface at the top and Bloc on the downward lope of the 60 incline. The tring pae over a frictionle, ale pulley. The incline are frictionle, but the horizontal urface ha coefficient of inetic friction 0.45. Bloc A and B have a 5g. The yte accelerate at /. ind the a of Bloc. SOLUTIO OTIUS BLOW: gin A gin ( A B a yte ow we need to find the friction force. or thi we need the oral force on B. ro the diagra we ee that B = B g. Since the bloc are in otion, we are dealing with inetic friction. g (0.45(5g 9.8 B ow we can ubtitute into the force forula:.05 ( 5g(9.8 in(60 (.05 (9.8 in(60 (5g 5g ( A bit of algebra will get u the deired a 64.5 (6.5 3g (8.5 84.5 0 ( or apu Learning Aitance Service at USB
6. A 0. g piece of wood i being held in place againt a vertical wall by a horizontal force of 5. ind the agnitude of the friction force acting on the wood. ro the force diagra, we can ee that the friction force ut equal the weight of the piece of wood to eep it fro falling. friction frictio n frictio n frictio n g (0.g(9.8.96 5 g or apu Learning Aitance Service at USB
7. A 4 g bloc on a horizontal urface i attached to a pring with a force contant of 50 /. A the pring and bloc are pulled forward at contant peed, the pring tretche by 5 c. ind the coefficient of inetic friction between the bloc and the table. The ey phrae here i contant peed. Since the bloc i oving at contant peed (and direction we now that it acceleration i 0. Thu the net force i 0 a well. Here i the force diagra. Since the net force i 0, we now that the oral force ut equal the weight, and the riction force ut equal the Spring force. friction pring The pring force obey Hooe Law: pring x The weight i: (50 g (4g(9.8 (0.5 39..5 inally, we can put thi together to find the friction: pring.5 0.3 frictio n (39. g or apu Learning Aitance Service at USB
50 8. A 000 g car i driven around a turn of radiu 50. What i the axiu afe peed of the car if the coefficient of tatic friction between the tire and the road i 0.75? The friction force ut be directed toward the center of the circle. Otherwie the car will lide off the road. If we want the axiu peed, then we want the axiu tatic friction force. The road i flat (not baned, o the oral force on the car i jut it weight. friction g (0.75(000g(9.8 friction,tatic,ax 7350 Thi friction force i the only force directed toward the center, o it ut be the centripetal force: v frictio n centripeta l r (000g(v 7350 50 v 9. or apu Learning Aitance Service at USB
9. A 90 g an drive hi car at a contant peed of 5 / over a all hill that ha a circular cro ection of radiu 40. ind hi apparent weight a he cret the top of the hill. (Hint: the apparent weight i the ae a the noral force on the an. 5 / When the car reache the top of the hill, it will have only force: it weight, and the noral force upplied by the road. Since the an i itting in the car, he feel a noral force a well. The car (and an ut be accelerating toward the center of the circle, o the net force on the an will be equal to the centripetal force required to eep hi oving along the circle. ce nt oral oral oral g g (90g(9.8 376 oral v r v r (90g(5 40 40 g or apu Learning Aitance Service at USB
0. Planet X ha a radiu that i 3 tie a large a and a a that i 6 tie that of. A ASA atronaut who weigh 550 here on i planning to ebar on a anned iion to Planet X. What will be the atronaut weight when he land there? The weight i the ae a the gravitational force. On the weight i 550: G M (r grav 550 In thi forula, G i contant, and will not change when the atronaut goe to a new planet. So the only part that will change are the a and radiu of the planet. The eay way to do thi proble i to iply ubtitute the new planet X value into the forula, and rearrange it: M r X X 3 gra v,x gra v,x gra v,x 6 r M G(6 M (3 r 6 3 G M 9 (r G M (r 3 (550 367 Planet X or apu Learning Aitance Service at USB
. Starting fro ret, a 0 g child decend a lide that i 4 high and ae an angle of 30 degree with the ground. At the botto of the lide the child ove with peed 3 /. What i the coefficient of inetic friction between the child and the lide? 4 A the child decend, potential energy i converted to inetic energy. riction alo doe oe wor againt the otion. We can ue conervation of energy to eep trac of the total energy. 30 ro the free-body diagra we ee that the noral force i gco(θ, which help find the friction force. Since the friction i parallel to the incline, we need to find the length of the incline to find the wor done by friction. botto v v (0g(3 algebra... 0.5 top gy gy W ( ( frictio n frictio n d gco( (0g(9.8 d (4 d=hypotenue of triangle ( (0g(9.8 co(30 4 ( in(30 or apu Learning Aitance Service at USB
. After tarting fro ret at a height of 40, a rollercoater pluet down a teep drop and then coplete a loop. eglecting friction, find the peed of the rollercoater when it i 0 eter below it tarting point. If the a of the rollercoater i 000 g, what ditance i required to top if a contant braing force of 8000 ewton i applied after the loop? y=0 y=0 y=40 Thi proble will ue conervation of energy for both part. The firt part i frictionle, o we jut have potential energy turning into inetic energy a the coater fall. v final Mv 9.8 initia l Mg(0 Mg(40 Brae applied here. d The econd part involve friction. Since the trac i horizontal where the friction i applied, the noral force on the coater i iply it weight. We are uing conervation of energy again. Thi tie all the energy i apped away by the friction force, o the wor done by friction hould equal the initial energy. 0 0 d final (000g(9.8 98 initia l Mg(40 W frictio n (40 (8000(d or apu Learning Aitance Service at USB
3. A bloc of a =4 g i itting on a table, preed againt a pring with force contant =00 /. Auing no friction, how far ut the pring be copreed o that the a will fly off the edge of the table at a peed of 5 /? Ue conervation of energy for thi one. The energy that i initially tored in the pring will all becoe inetic energy when the bloc leave the pring. x final v (4g(5 initia l x (00 x x 5 / IITIAL IAL or apu Learning Aitance Service at USB