HW11.sol - homework 11 – RAHMAN TARIQUE – Due Mar 3 2008 3:00 am 1 Question 1 chap 8 sect 1 part 1 of 1 10 points A weight lifter lifts a mass

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Unformatted text preview: homework 11 – RAHMAN, TARIQUE – Due: Mar 3 2008, 3:00 am 1 Question 1, chap 8, sect 1. part 1 of 1 10 points A weight lifter lifts a mass m at constant speed to a height h in time t . How much work W is done by the weight lifter? 1. W = mg h t 2. W = mg ht 3. W = mg 4. W = mg h correct 5. W = mh Explanation: By conservation of energy, the work done to the mass is the change in its energy. Since Δ V = 0, only the gravitational potential en- ergy changes, so the work is W = Δ U = mg h . Question 2, chap 8, sect 1. part 1 of 1 10 points A frictionless pendulum of length 3 m swings with an amplitude of 10 ◦ . At its max- imum displacement, the potential energy of the pendulum is 10 J. What is the kinetic energy of the pendulum when its potential energy is 5 J? 1. K = 5 J correct 2. K = 3 . 3 J 3. K = 15 J 4. K = 6 . 7 J 5. K = 10 J Explanation: The total energy is conserved, which means the sum of the potential energy and the ki- netic energy is a constant. At the maximum displacement, the potential energy of the pen- dulum is E = 10 J. Assuming at the lowest point the potential energy of the pendulum is zero, we know the sum of the energies is 10 J. So if the potential energy is U = 5 J, the kinetic energy is K = 10 J − 5 J = 5 J. Question 3, chap 8, sect 1. part 1 of 1 10 points A block of mass m slides on a horizontal frictionless table with an initial speed v . It then compresses a spring of force constant k and is brought to rest. The acceleration of gravity is 9 . 8 m / s 2 . v m k m μ = 0 How much is the spring compressed x from its natural length? 1. x = v radicalbigg mg k 2. x = v 2 2 m 3. x = v radicalbigg m k correct 4. x = v 2 2 g 5. x = v mk g 6. x = v m k g 7. x = v k g m 8. x = v radicalBigg k mg 9. x = v mg k 10. x = v radicalbigg k m Explanation: Total energy is conserved (no friction). The spring is compressed by a distance x from its homework 11 – RAHMAN, TARIQUE – Due: Mar 3 2008, 3:00 am 2 natural length, so 1 2 mv 2 = E i = E f = 1 2 k x 2 , or x 2 = m k v 2 , therefore x = v radicalbigg m k . Anyone who checks to see if the units are correct should get this problem correct. Question 4, chap 8, sect 1. part 1 of 1 10 points A 5 kg mass is attached to a light cord that passes over a massless, frictionless pulley. The other end of the cord is attached to a 3 . 2 kg mass. The acceleration of gravity is 9 . 8 m / s 2 . 7 . 8 m ω 5 kg 3 . 2 kg Use conservation of energy to determine the final speed of the first mass after it has fallen (starting from rest) 7 . 8 m. Correct answer: 5 . 79302 m / s (tolerance ± 1 %). Explanation: Let : m 1 = 5 kg , m 2 = 3 . 2 kg , and ℓ = 7 . 8 m . Consider the free body diagrams 5 kg 3 . 2 kg T T m 1 g m 2 g a a Let the figure represent the initial config- uration of the pulley system (before m 1 falls down)....
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This note was uploaded on 02/03/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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HW11.sol - homework 11 – RAHMAN TARIQUE – Due Mar 3 2008 3:00 am 1 Question 1 chap 8 sect 1 part 1 of 1 10 points A weight lifter lifts a mass

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