Kaplunovsky - hw 05 - homework 05 Due: Feb 22 2007, 4:00 am...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. Newtons 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 astronauts 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 , = 21 , and g = 9 . 8 m / s 2 . T T sin T cos mg Vertically summationdisplay F y = T cos mg = 0 T cos = mg . (1) Horizontally, summationdisplay F x = T sin = ma . (2) Dividing Eqs 1 and 2, we have T sin T cos = a g tan = a g a = g tan = ( 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....
View Full Document

Page1 / 13

Kaplunovsky - hw 05 - homework 05 Due: Feb 22 2007, 4:00 am...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online