p201f04-homework07-sol

p201f04-homework07-sol - Version One Homework 7 Heinz 81204...

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: Version One Homework 7 Heinz 81204 Oct 07, 2004 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. Serway CP 05 76 08:05, trigonometry, numeric, > 1 min, wording-variable. 001 (part 1 of 1) 10 points A 2 kg block situated on a rough incline with 37 angle is connected to a spring of negligible mass having a spring constant of 100 N / m. The block is released from rest when the spring is unstretched, and the pulley is fric- tionless. The block moves 20 cm down the incline before coming to rest. The acceleration of gravity is 9 . 8 m / s 2 . 100 N / m 2 k g 2 c m 37 Find the coefficient of kinetic friction be- tween block and incline. Correct answer: 0 . 114709 . Explanation: Given : m = 2 kg , = 37 , k = 100 N / m , and x = 20 cm . The work done by the friction force is W f = ( f x cos180 ) x =- k n x =- k ( mg cos ) x Applying the work-kinetic energy theorem, W nc = W f = K + U s + U g . Since the block stars from rest and comes to a stop,- mg x cos = 0 + 1 2 k ( x ) 2- mg x sin . Thus =- k x 2 mg cos + tan =- (100 N / m)(0 . 2 m) 2(2 kg)(9 . 8 m / s 2 )cos37 + tan37 =- (0 . 638845) + (0 . 753554) = . 114709 . Return to the Shuttle 09:01, trigonometry, numeric, > 1 min, nor- mal. 002 (part 1 of 1) 10 points A(n) 65 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 50 m away from the shuttle and mov- ing with zero speed relative to the shuttle. She has a(n) 0 . 65 kg camera in her hand and decides to get back to the shuttle by throw- ing the camera at a speed of 12 m / s in the direction away from the shuttle. How long will it take for her to reach the shuttle? Answer in minutes. Correct answer: 6 . 94446 min. Explanation: Let : M = 65 kg , m = 0 . 65 kg , d = 50 m , and v = 12 m / s . Because of conservation of linear momentum, we have 0 = M V + m (- v ) mv = M V Version One Homework 7 Heinz 81204 Oct 07, 2004 2 where V is the velocity of the astronaut and it has a direction toward the shuttle. V = mv M = (0 . 65 kg)(12 m / s) 65 kg = 0 . 12 m / s . And the time it takes for her to reach the shuttle t = d V = 50 m . 12 m / s 1 min 60 s = 6 . 94446 min . Serway CP 06 59 09:01, trigonometry, numeric, > 1 min, nor- mal. 003 (part 1 of 2) 5 points A small block of mass 55 g is released from rest at the top of a curved frictionless wedge of mass 333 g which sits on a frictionless hori- zontal surface as shown....
View Full Document

This note was uploaded on 07/02/2011 for the course PHYS 201 taught by Professor Woodahl during the Fall '09 term at Indiana.

Page1 / 7

p201f04-homework07-sol - Version One Homework 7 Heinz 81204...

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