{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW14_Solutions

# HW14_Solutions - hiarker(srh959 homework 14 Turner(58220...

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

hiarker (srh959) – homework 14 – Turner – (58220) 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 5 . 89 m on the incline by a 150 N force. The acceleration of gravity is 9 . 8 m / s 2 . 13 kg μ = 0 . 303 150 N 1 . 48 m / s 22 a) What is the change in kinetic energy of the crate? Correct answer: 391 . 589 J. Explanation: Let : F = 150 N , d = 5 . 89 m , θ = 22 , m = 13 kg , g = 9 . 8 m / s 2 , μ = 0 . 303 , and v = 1 . 48 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 . 303) (13 kg) (9 . 8 m / s 2 ) × (5 . 89 m) cos 22 = 210 . 811 J . The work due to the applied force is W appl = F d = (150 N) (5 . 89 m) = 883 . 5 J , and the work due to gravity is W grav = m g d sin θ = (13 kg) (9 . 8 m / s 2 ) × (5 . 89 m) sin 22 = 281 . 1 J , so that Δ K = W fric + W appl + W grav = ( 210 . 811 J) + (883 . 5 J) + ( 281 . 1 J) = 391 . 589 J . 002 (part 2 of 2) 10.0 points b) What is the speed of the crate after it is pulled the 5 . 89 m? Correct answer: 7 . 90158 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

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

View Full Document
hiarker (srh959) – homework 14 – Turner – (58220) 2 v f = radicalbigg 2 Δ K m + v 2 i = radicalBigg 2(391 . 589 J) 13 kg + (1 . 48 m / s) 2 = 7 . 90158 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. 592 g 5 . 1 m 2 . 1 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: 7 . 67203 m / s. Explanation: Let : g = 9 . 81 m / s 2 , m = 592 g , and h 1 = 3 m . m h h 1 h 2 5 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 ) (3 m) = 7 . 67203 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 ]}

### Page1 / 6

HW14_Solutions - hiarker(srh959 homework 14 Turner(58220...

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

View Full Document
Ask a homework question - tutors are online