Physics 6A. Practice Midterm #2 solutions

Transcription

1 Phyic 6A Practice Midter # olution

2 1. 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 think about thi one i that ince the a i cut in half, the acceleration ut double.

3 . Two boxe are placed next to each other on a ooth flat urface. Box A ha a 1 kg and Box B ha a 3 kg. A contant horizontal force of 8 i applied to Box A. ind the force exerted on Box B. We can think of thi a a ingle box with total a 4kg. Then uing a we get the acceleration of the whole yte. 8 a 4 yte kg a yte ow we do the ae thing, but jut for box B: A B B 3kg B 6 Reeber Box B ha a a of only 3kg

4 3. Two boxe are placed next to each other on a flat urface. Box A ha a 1kg and Box B ha a 3kg. The coefficient of friction are 0.3 and 0.4 for kinetic and tatic friction, repectively. A contant horizontal force of 8 i applied. ind the acceleration of Box B. We can think of thi a a ingle box with total a 4kg. 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 μ tatic,ax ( ) ( ) 0.4 4kg 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 keep the fro oving.

5 4. A crate filled with book ret on a horizontal floor. The total weight of the crate and book i 700. The coefficient of friction are 0.35 for kinetic 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: x y co in o ( 0 ) 43 o ( 0 ) 154 friction 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 Thi give a axiu force of tatic friction (0.45)(854) 384 y 700 oral 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 kinetic. friction,k (0.35)(854) 99 ow we can ue a again for the x-direction force: 700 net,x x friction,k a a a x

6 5. Block A and B and are connected by a ale tring and placed a hown, with Block A on the upward lope of the 60 incline, Block B on the horizontal urface at the top and Block 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 kinetic friction Block A and B have a 5kg. The yte accelerate at /. ind the a of Block. k B B T a / T 1 T 1 B g T A A A gcoθ gcoθ ginθ A ginθ A g g Here i the force diagra for thi proble. The block 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: Aginθ k + ginθ ( A + B + ) ayte 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 1 on A i canceled by the backward tenion T 1 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.

7 5. Block A and B and are connected by a ale tring and placed a hown, with Block A on the upward lope of the 60 incline, Block B on the horizontal urface at the top and Block 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 kinetic friction Block A and B have a 5kg. The yte accelerate at /. ind the a of Block. A ginθ k + ginθ ( A + B + ) ayte 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 block are in otion, we are dealing with kinetic friction. ( 0.45)( 5kg 9.8 ).05 μk Bg k ow we can ubtitute into the force forula: ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) o o 5kg 9.8 in in 60 5kg + 5kg + A bit of algebra will get u the deired a ( ) kg 84.5 ( 8.5) 0 + ( )

8 6. A 0. kg 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 keep it fro falling. friction friction friction friction g ( 0.kg)( 9.8 ) g

9 7. A 4 kg block on a horizontal urface i attached to a pring with a force contant of 50 /. A the pring and block are pulled forward at contant peed, the pring tretche by 5 c. ind the coefficient of kinetic friction between the block and the table. The key phrae here i contant peed. Since the block i oving at contant peed (and direction) we know 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 know that the oral force ut equal the weight, and the riction force ut equal the Spring force. friction pring The pring force obey Hooke Law: The weight i: ( 50 )( 0.5) 1.5 pring k x ( 4kg)( 9.8 ) 39. g inally, we can put thi together to find the friction: 1.5 μ μ pring k 0.3 friction k μ k ( 39.) g

10 8. A 1000 kg 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 banked), o the oral force on the car i jut it weight. friction 50 ( 0.75)( 1000kg)( 9.8 ) 7350,tatic,ax μ g friction Thi friction force i the only force directed toward the center, o it ut be the centripetal force: friction 7350 v 19. ( 1000kg)( v ) centripetal 50 v r

11 9. A 90 kg an drive hi car at a contant peed of 15 / 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.) 15 / 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 keep hi oving along the circle. v cent g oral r v oral g r oral oral 376 ( 90kg ( )( ) )( 15 ) 90kg g

12 10. Planet X ha a radiu that i 3 tie a large a Earth and a a that i 6 tie that of Earth. A ASA atronaut who weigh 550 here on Earth i planning to ebark 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 Earth the weight i 550: G M Earth grav 550 ( rearth) 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: MX 6 MEarth rx 3 rearth G( 6 M ) Earth grav,x ( 3 rearth) 6 G M Earth grav,x 9 ( rearth) G M Earth grav,x { 550} ( r ) 3 Earth Earth Planet X