# hw16 - hinojosa(jlh3938 homework 16 Turner(58185 This...

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

hinojosa (jlh3938) – homework 16 – Turner – (58185) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points 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 7 . 94 m on the incline by a 150 N force. The acceleration of gravity is 9 . 8 m / s 2 . 13 kg μ = 0 . 308 150 N 1 . 26 m / s 35 a) What is the change in kinetic energy of the crate? Correct answer: 355 . 581 J. Explanation: Let : F = 150 N , d = 7 . 94 m , θ = 35 , m = 13 kg , g = 9 . 8 m / s 2 , μ = 0 . 308 , and v = 1 . 26 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) (13 kg) (9 . 8 m / s 2 ) × (7 . 94 m) cos 35 = 255 . 214 J . The work due to the applied force is W appl = F d = (150 N) (7 . 94 m) = 1191 J , and the work due to gravity is W grav = m g d sin θ = (13 kg) (9 . 8 m / s 2 ) × (7 . 94 m) sin 35 = 580 . 205 J , so that Δ K = W fric + W appl + W grav = ( 255 . 214 J) + (1191 J) + ( 580 . 205 J) = 355 . 581 J .

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

View Full Document
hinojosa (jlh3938) – homework 16 – Turner – (58185) 2 002 (part 2 of 2) 10.0 points b) What is the speed of the crate after it is pulled the 7 . 94 m? Correct answer: 7 . 50282 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 v f = radicalbigg 2 Δ K m + v 2 i = radicalBigg 2(355 . 581 J) 13 kg + (1 . 26 m / s) 2 = 7 . 50282 m / s . 003 (part 1 of 3) 10.0 points 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. 412 g 4 . 1 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 . 94273 m / s. Explanation: Let : g = 9 . 81 m / s 2 , m = 412 g , and h 1 = 1 . 8 m . m h h 1 h 2 4 . 1 m 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 . 8 m) = 5 . 94273 m / s . 004 (part 2 of 3) 10.0 points What horizontal distance does the block travel in the air?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern