PHY302K-Sai-Midterm 2-solutions

PHY302K-Sai-Midterm 2-solutions - Version 120 Midterm 2...

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Version 120 – Midterm 2 – sai – (58015) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points According to the work-energy theorem, iF work is done on an object, its potential and/or kinetic energy changes. Consider a car that accelerates From rest on a ±at road. What Force did the work that increased the car’s kinetic energy? 1. gravity 2. the Force oF the car engine 3. the Friction between the road and the tires correct 4. air resistance Explanation: The Force oF Friction between the road and the tires pushes Forward on the car and does the work that increased the car’s kinetic en- ergy. 002 10.0 points A block with mass m starting From rest at point A is sliding down a rough incline with kinetic coe²cient oF Friction μ . The incline has angle θ with respect to the horizontal sur- Face. As the block slides down For a distance s , it passes point B . m s μ A B v =0 θ ³ind the kinetic energy oF the block as it passes B . 1. K B = mg s cos θ + μmg s sin θ 2. K B = mg s sin θ - μmg s sin θ 3. K B = mg s sin θ - μmg s cos θ correct 4. K B = mg s cos θ - μmg s cos θ 5. K B = mg s cos θ - μmg s sin θ 6. K B = mg s cos θ + μmg s cos θ 7. K B = mg s sin θ + μmg s sin θ 8. K B = mg s sin θ + μmg s cos θ Explanation: The Friction Force is given by f = μ N = μmg cos θ , so the work done by the Friction Force is W f = - fs = - μmg s cos θ . (1) mg cos f sin N mg We know E B - E A =( K B + P B ) - ( K A + P A ) =( K B + 0) - (0 + mg h ) = K B - mg s sin θ (2) Since E B - E A = W f by the work energy theorem, and using Eqs. (1) and (2) we have K B = mg s sin θ - μmg s cos θ . 003 10.0 points
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Version 120 – Midterm 2 – sai – (58015) 2 An ore car of mass 36000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 21 m lower vertically, is a horizontally situated spring with constant 3 . 3 × 10 5 N / m. The acceleration of gravity is 9 . 8m / s 2 . Ignore friction. How much is the spring compressed in stop- ping the ore car? 1. 6.70088 2. 7.19333 3. 5.06539 4. 5.84556 5. 7.93035 6. 5.34743 7. 7.05886 8. 7.7106 9. 8.78958 10. 5.00213 Correct answer: 6 . 70088 m. Explanation: Energy is conserved, so the change of po- tential energy from when the car is at rest to when it just hits the spring is mg h = 1 2 mv 2 . The kinetic energy is then converted to po- tential energy in the spring as the cart comes to rest. When the spring is fully compressed by an amount d , all of the kinetic energy has been converted to potential energy so 1 2 mv 2 = 1 2 kd 2 . Thus, 1 2 kd 2 = mg h, and solving for d we have d = ± 2 mg h k = ² 2 (36000 kg) (9 . 8m / s 2 ) (21 m) (3 . 3 × 10 5 N / m) = 6 . 70088 m .
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This note was uploaded on 02/22/2012 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas.

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PHY302K-Sai-Midterm 2-solutions - Version 120 Midterm 2...

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