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Kaplunovsky - hw 05

# Kaplunovsky - hw 05 - homework 05 – – Due 4:00 am 1...

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Unformatted text preview: homework 05 – – Due: Feb 22 2007, 4:00 am 1 Question 1, chap 5, sect 2. part 1 of 1 6 points An astronaut who weighs 554 . 9 N on the surface of the earth lifts off from planet Zuton in a space ship. The free-fall acceleration on Zuton is 5 . 8 m / s 2 . At the moment of lift off the space ship experiences an acceleration of 7 . 9 m / s 2 . The acceleration of gravity is 9 . 8 m / s 2 . What is the magnitude of the force that the space ship exerts on the astronaut? Correct answer: 775 . 728 N (tolerance ± 1 %). Explanation: There are two forces acting on the astro- naut: The spaceship pushing him upward and gravity pulling him down. Newton’s Second Law relates his acceleration to the net force, so ma = F net = F ship − mg Z where g Z is the free-fall acceleration on Zuton. Consequently, F ship = m ( a + g Z ) . It remains to calculate the astronaut’s mass in terms of his weight on Earth , m = W on Earth g Earth = 554 . 9 N 9 . 8 m / s 2 = 56 . 6224 kg . Therefore, F ship = (56 . 6224 kg)(7 . 9 m / s 2 + 5 . 8 m / s 2 ) = 775 . 728 N . Question 2, chap 5, sect 6. part 1 of 2 6 points An elevator accelerates upward at 1 . 2 m / s 2 . The acceleration of gravity is 9 . 8 m / s 2 . What is the upward force exerted by the floor of the elevator on a(n) 79 kg passenger? Correct answer: 869 N (tolerance ± 1 %). Explanation: N a mg When the elevator is accelerating upward, F net = ma = N − mg N = ma + mg. Question 3, chap 5, sect 6. part 2 of 2 6 points If the same elevator accelerates downwards with an acceleration of 1 . 2 m / s 2 , what is the upward force exerted by the elevator floor on the passenger? Correct answer: 679 . 4 N (tolerance ± 1 %). Explanation: N a mg When the elevator is accelerating down- ward, F net = ma = mg − N 2 N 2 = mg − ma. Question 4, chap 5, sect 6. part 1 of 1 8 points A 2 . 2 kg object hangs at one end of a rope that is attached to a support on a railroad boxcar. When the car accelerates to the right, the rope makes an angle of 21 ◦ with the verti- cal The acceleration of gravity is 9 . 8 m / s 2 . a 2 . 2 kg 2 1 ◦ Find the acceleration of the car. ( Hint: vectora object = vectora car ) homework 05 – – Due: Feb 22 2007, 4:00 am 2 Correct answer: 3 . 76187 m / s 2 (tolerance ± 1 %). Explanation: Given : m = 2 . 2 kg , θ = 21 ◦ , and g = 9 . 8 m / s 2 . T T sin θ T cos θ θ mg Vertically summationdisplay F y = T cos θ − mg = 0 T cos θ = mg . (1) Horizontally, summationdisplay F x = T sin θ = ma . (2) Dividing Eqs 1 and 2, we have T sin θ T cos θ = a g tan θ = a g a = g tan θ = ( 9 . 8 m / s 2 ) tan21 ◦ = 3 . 76187 m / s 2 . Question 5, chap 5, sect 4. part 1 of 1 8 points One of the great dangers to mountain climbers is an avalanche, in which a large mass of snow and ice breaks loose and goes on an essentially frictionless “ride” down a mountainside on a cushion of compressed air....
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Kaplunovsky - hw 05 - homework 05 – – Due 4:00 am 1...

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