Solutions to Quiz 4B in Physics 2050B, Fall 2008 - Prof. Lisa M. Paulius, Quizzes of Physics

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Quiz 4B Solutions Physics 2050B Name: Fall 2008 Please print your name clearly You must show ALL your work to receive credit. Include units on your answers. Please circle your answer. g = 9.8 m/s^2

1. Two masses are connected by a rope going over a massless fritionless pulley as shown. MA = 10 kg, MB = 30 kg. Which of the following statements is true? Part 1 (2 pts) a) TA >TB b) TA=TB the only effect of a massless, frictionless pulley is to change the direction of the force, so the tension is the same everywhere in the one rope c) TA<TB Part 2 (2 pts) a) TA>MA g we know the masses will start accelerating in the direction of the larger weight, so Mass A will start accelerating upward and Mass B will accelerate downward. In order for this to happen, the weight of A must be less than the tension resisting its falling. b) TA= MA g c) TA< MA g
2. (3 pts) A block starts with an initial velocity up a rough incline. Which graph best represents the block’s velocity as a function of time? QuickTime™ and a decompressor are needed to see this picture. MA M B TA TB QuickTime™ and a decompressor are needed to see this picture. are needed to see this picture.^ QuickTime™ and adecompressor

2 (18 pts) A child is dragging her sled across a rough surface with constant velocity by pulling on a rope with a force F with an angle of 40° as shown in the diagram. The mass of the sled is 15 kg. a) (6 pts) Draw the free body diagram for the sled. b) (12 pts) Determine the Force F if the coefficient of friction between the sled and the ground is 0.54. We drew the free body diagram, Now we break the forces into components along x &y directions Then we use Newton’s 2nd^ Law in both directions In the y direction: €

∑^ Fy =^ n^ +^ F^ sinΘ^ −^ mg^ =^ may

since the sled will not lift off the surface or sink down into it ay= or €

∑^ Fy =^ n^ +^ F^ sinΘ^ −^ mg^ =^0 ⇒

n = mg − F sinΘ In the x direction €

∑^ Fx =^ F^ cosΘ^ −^ f^ =^ max

Since the velocity is constant => ax= And we can rewrite f as μn So we get €

∑^ Fx =^ F^ cosΘ^ −μ n^ =^0

plug in n from the solutions in the y direction and we get €

∑^ Fx =^ F^ cosΘ^ −^ μ( mg^ −^ F^ sin θ )^0 ⇒^ )

F cosΘ + μ F sin θ = μ mg ⇒ F (cosΘ + μ sin θ ) = μ mg ⇒ F = μ mg (cosΘ + μ sin θ ) F = ( 0. 54 )( 15 kg )( 9. 8 m / s^2 ) (cos 40 ° + 0. 54 sin 40 °) F =71.31N Q Q F mg f n

Problem 1 (20 pts) A small rocket with a mass of 300 kg floats in outer space and has two different jets firing at the same time with constant forces acting in the directions shown. FA = 250 N, QA = 15° , FB = 600 N, QB = 37° a) (12 pts) What acceleration will the rocket experience? We need to find the sum of the forces, so add the two vectors by breaking them into components Fx Fy FA =-FAsin(15) =FAcos(15) -64.702 241. FB =FB cos(37) =-FBsin(37) 479.2N5 -361.1N Fnet 414.5N5 -119.6N The vector a is the vector Fne t/m So in component form a = F /m = 1.38m/ s^2 i

• -0.40m/ s^2 j Or the magnitude is anet = sqrt(ax^2 + ay^2 ) |a|=1.43m/s^2 direction can be found from tanQ=ay/ax=> Q b) (4 pts) How far will it have moved in the first 7 s (assuming it starts from rest)? Quickest is to use the full acceleration d=1/2at 2 =(0.5)*(1.43m/s 2 )(7s) 2 =35.23m (Note of caution here – this only works because there was not an initial velocity, If there had been we would have had to find x and y separately) c) (4 pts) In what direction will it have moved? It will move along the direction of the net acceleration (which is || to the net force) which we found above Q FA FB QA QB

Problem 3 (20 pts) A 5 kg block starts at the bottom of a ramp with an initial speed of 20 m/s up the ramp. If the coefficient of friction between the block and the ramp is μ=0.6, the angle Q is 60°, a) (6 pts) Determine the magnitude of the normal force acting on the block First we draw the free body diagram (forces shown in blue) Then we break the forces into components (components of mg shown in green, other forces already || or  to plane) Using || to plane as my x directinon and perpendicular to as my y direction We can use Newton’s 2 nd Law in the direction  to plane SFy=n-mgcosQ=may, there is no acceleration  to plane so ay= => n=mgcosQ n=(5kg)(9.8m/s^2 )cos60° n=24.5N b) (14 pts) determine the distance d up the ramp the block slides before coming to a stop. We can use Newton’s 2 nd Law in the x direction (|| to plane) SFx=-f-mgsinQ=max, SFx=-μn-mgsinQ=max  (^) a =(-μn-mgsinQm  (^) a =[(-(0.6)(24.5N)-(5kg)(9.8m/s 2 )(sinkg)=-m/s^2 a is constant, so we can use the kinematic equations then the quickest solution would be to use equation 3 with vf=0, vo=20m/s and a =-11.42m/ s 2 vx^2 - vxo^2 = 2ax(x-xo) (x-xo)=-vxo^2 /2ax d=(x-xo)=-(8m/s) 2 /(2)*(-11.42m/s 2 )) d=17.5m Q  Q      Q  Q  Q