{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ohw 14 - brown(twb493 ohw 14 turner(57340 This print-out...

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

View Full Document Right Arrow Icon
brown (twb493) – ohw 14 – turner – (57340) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points A crate is pulled by a force (parallel to the incline) up a rough incline. The crate has an initial speed shown in the figure below. The crate is pulled a distance of 8 . 01 m on the incline by a 150 N force. The acceleration of gravity is 9 . 8 m / s 2 . 12 kg μ = 0 . 308 150 N 1 . 71 m / s 29 a) What is the change in kinetic energy of the crate? Correct answer: 491 . 069 J. Explanation: Let : F = 150 N , d = 8 . 01 m , θ = 29 , m = 12 kg , g = 9 . 8 m / s 2 , μ = 0 . 308 , and v = 1 . 71 m / s . F μ N N m g v θ The work-energy theorem with nonconser- vative forces reads W fric + W appl + W gravity = Δ K To find the work done by friction we need the normal force on the block from Newton’s law summationdisplay F y = N − m g cos θ = 0 ⇒ N = m g cos θ . Thus W fric = μ m g d cos θ = (0 . 308) (12 kg) (9 . 8 m / s 2 ) × (8 . 01 m) cos 29 = 253 . 752 J . The work due to the applied force is W appl = F d = (150 N) (8 . 01 m) = 1201 . 5 J , and the work due to gravity is W grav = m g d sin θ = (12 kg) (9 . 8 m / s 2 ) × (8 . 01 m) sin 29 = 456 . 679 J , so that Δ K = W fric + W appl + W grav = ( 253 . 752 J) + (1201 . 5 J) + ( 456 . 679 J) = 491 . 069 J . 002(part2of2)10.0points b) What is the speed of the crate after it is pulled the 8 . 01 m? Correct answer: 9 . 207 m / s. Explanation: Since 1 2 m ( v 2 f v 2 i ) = Δ K v 2 f v 2 i = 2 Δ K m
Image of page 1

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

View Full Document Right Arrow Icon
brown (twb493) – ohw 14 – turner – (57340) 2 v f = radicalbigg 2 Δ K m + v 2 i = radicalBigg 2(491 . 069 J) 12 kg + (1 . 71 m / s) 2 = 9 . 207 m / s . 003(part1of3)10.0points A block starts at rest and slides down a fric- tionless track. It leaves the track horizontally, flies through the air, and subsequently strikes the ground. 408 g 3 . 7 m 2 . 3 m x v What is the speed of the ball when it leaves the track? The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 5 . 24099 m / s. Explanation: Let : g = 9 . 81 m / s 2 , m = 408 g , and h 1 = 1 . 4 m . m h h 1 h 2 x v Choose the point where the block leaves the track as the origin of the coordinate system. While on the ramp, K b = U t 1 2 m v 2 x = m g h 1 v 2 x = 2 g h 1 v x = radicalbig 2 g h 1 = radicalBig 2 ( 9 . 81 m / s 2 ) (1 . 4 m) = 5 . 24099 m / s . 004(part2of3)10.0points What horizontal distance does the block travel in the air? Correct answer: 3 . 58887 m. Explanation: Let : h 2 = 2 . 3 m . With the point of launch as the origin, h 2 = 1 2 g t 2 t = radicalBigg 2 h 2 g . Thus x = v x t = v x radicalBigg 2 h 2 g = (5 . 24099 m / s) radicalBigg 2 ( 2 . 3 m) 9 . 81 m / s 2 = 3 . 58887 m . 005(part3of3)10.0points What is the speed of the block when it hits the ground?
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern