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PHYS_2014_Sample_Final_Exam_soln-1

PHYS_2014_Sample_Final_Exam_soln-1 - PHYS2014/Fall...

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PHYS2014/Fall 2010/Sample Final Exam Name:___________________________ OSU Physics Dept./Benton Recitation TA and Time:_________________ 1 Physics 2014: General Physics I Solutions to Sample Final Exam 1. While driving down a hill of constant slope θ = 5 ° in a 1600 kg car at a speed of 25 m/s, you slam on the brakes such that the wheels immediately lock up and the car is sliding on the pavement. The coefficient of friction, μ k , between the tires and pavement is 0.70. How far does the car skid before coming to a stop? Solution: a) We can solve this kinematically or using Conservation of Energy. Let’s first solve it using conservation of energy. Conservation of energy tells us that the gravitational potential energy of the car plus the kinetic energy of the car (initial) will be completely converted into frictional or thermal energy (final). 2 f 1 2 g th U K U mgh mv F s + = + = Δ The energy lost to friction will be the force of friction applied over a distance, Δ s, which is the quantity we want to find. f cos cos sin k N k th k F F mg U mg s h s m μ μ θ μ θ θ = = = Δ = Δ 1 sin 2 g s m θ Δ + 2 k v m μ = ( ) ( ) ( ) ( ) 2 2 2 2 cos 1 cos sin cos sin 2 25m s 52.3m 2 cos sin 2 9.8m s 0.70 cos5 sin5 k k k g s v g s g s sg v s g θ μ θ θ μ θ θ μ θ θ Δ = Δ Δ = Δ Δ = = = ringoperator ringoperator We can also solve the problem using forces and kinematics. In solving the problem this way, it is probably best to rotate the coordinate system by 5 ° clockwise. θ = 5 ο
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PHYS2014/Fall 2010/Sample Final Exam Name:___________________________ OSU Physics Dept./Benton Recitation TA and Time:_________________ 2 f f 0 cos cos sin cos y N N x k N k k F F mg F mg F F mg ma F F mg m θ θ θ μ μ θ μ = = = = − + = = = cos g m θ + sin g m θ = ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 sin cos m m 9.80 sin5 0.70 cos5 5.98 s s 2 0 m 25 s 52.3m 2 m 2 5.98 s k f i f i a a g a v v a s v v s a θ μ θ = = = − = + Δ = Δ = = = ⋅ − ringoperator ringoperator f F arrowrightnosp N F arrowrightnosp g F arrowrightnosp θ = 5 ο g F arrowrightnosp gx F arrowrightnosp gy F arrowrightnosp
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PHYS2014/Fall 2010/Sample Final Exam Name:___________________________ OSU Physics Dept./Benton Recitation TA and Time:_________________ 3 2. Aliens from the planet Splorg have established an orbital base above the Earth from which to launch missions to create crop circles, tip cows, and abduct humans for medical examinations. The CIA, working in close collaboration with scientists from NASA, have estimated that the distance of the Splorigian base in its circular orbit is 7.66 × 10 8 m from the center of the Earth and that the mass of the Splorigian base is 38,000 kg. a) What is the orbital period of the Splorigian base? b) What is the kinetic energy of the Splorigian base in this orbit? Solution: a) The centripetal force required to keep the Splorigian base in orbit is provided by the Earth’s gravity: 2 2 2 2 4 Mm m r G r T m ω π ω π = = 2 r Mm G T = ( ) 2 2 3 2 3 2 8 2 3 2 11 24 2 6 4 4 7.66 10 m 4 m 6.67 10 N 5.98 10 kg kg 6.67 10 s or 77.2 days r r T GM r T GM T π π π = × = = × × = × b) The kinetic energy for a spacecraft in a circular orbit is given by: 1 2 24 2 11 17 2 8 2 5.98 10 kg 38,000kg m 6.67 10 N 9.89 10 J kg 2 7.66 10 m mm K G r K = × = × = × ×
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