f5ch05 - CHAPTER 5 FORCE AND MOTION I 113 CHAPTER 5 Answers...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 5 FORCE AND MOTION – I 113 C HAPTER 5 Answers to Checkpoint Questions . c, d, and e . ( a ) and ( b ) N, leftward (acceleration is zero in each situation . ( a ) and ( b ) , , , . ( a ) equal; ( b ) greater (acceleration is upward, thus net force on body must be upward) . ( a ) equal; ( b ) greater; ( c ) less . ( a ) increase; ( b ) yes; ( c ) same; ( d ) yes . ( a ) F sin ; ( b ) increase . Answers to Questions . ( a ) yes; ( b ) yes; ( c ) yes; ( d ) yes . yes: B , C , and D . ( a ) and ; ( b ) and . ( a ) increases from an initial value of mg ; ( b ) decreases from mg to zero (after which the block moves up away from the oor) . ( a ) N, upward; ( b ) N, upward . ( a ) supports; ( b ) less; ( c ) F sin . ( a ) less; ( b ) greater . ( a ) when stationary and when moving (up or down) at constant speed; ( b ) when accelerating upward from the ground oor and when stopping back at the ground oor; ( c ) when stopping at the top oor and when accelerating downward from the top oor . ( a ) no; ( b ) no; ( c ) no . ( a ) W ; ( b ) W ; ( c ) W ; ( d ) W
Background image of page 1

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

View Full DocumentRight Arrow Icon
114 CHAPTER 5 FORCE AND MOTION – I . ( a ) increases; ( b ) increases; ( c ) decreases; ( d ) decreases . ( a ) equal; ( b ) equal; ( c ) equal; ( d ) less; ( e ) greater . ( a ) kg; ( b ) kg; ( c ) kg; ( d ) all tie; ( e ) , , . ( a ) kg; ( b ) kg; ( c ) kg; ( d ) all tie; ( e ) F , F , F . d, c, a, b E ( a ) Using F = m a , we nd the x component of the force to be F x = ma x = ma cos = ( : kg)( : m/s )(cos ) = : N and the y component of the force to be F y = ma y = ma sin = ( : kg)( : m/s )(sin ) = : N : ( b ) In unit-vector notation, the force vector is F = F x i + F y j = ( : i + : j ) N : E ( a ) The net force is F net = F + F = ( : N) i + ( : N) j + ( : N) i + ( : N) j = ( : i : j ) N : ( b ) The magnitude of F net is j F net j = p ( : N) + ( : N) = : N ; and the angle which F net makes with the positive x axis is given by = tan F net ;y F net ;x = : N : N = : ( c ) Use a = F net =m to compute a = j a j : a = F net m = : N : kg = : m/s :
Background image of page 2
CHAPTER 5 FORCE AND MOTION – I 115 The direction of a is the same as that of F net : P ( a ) Let the other force be F = F x i + F y j . Then the net force acting on the body is F net = F + F . Using F net = m a , we have F x + F x = : N + F x = ma x = ma cos = ( kg)( : m/s )(cos ) in the x direction and F y + F y = : N + F y = ma y = ma sin = ( kg)( : m/s )(sin ) in the y direction. Solve for F x and F y to get F x = : N and F y = : N : Thus F = F x i + F y j = ( : i : j ) N : ( b ) The magnitude of F is given by F = q F x + F y = p ( : ) + ( : ) N = : N : The vector F makes an angle with the positive x axis, where = tan F y F x = tan : N : N : The value of satisfying the above equality is either = : or = . Since F x < and F y < , we reject the rst solution and obtain = : E The net force applied on the choping block is F net = F + F . Thus the acceleration of the block is given by a = ( F + F ) =m . ( a
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 29

f5ch05 - CHAPTER 5 FORCE AND MOTION I 113 CHAPTER 5 Answers...

This preview shows document pages 1 - 4. Sign up to view the full document.

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