{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# SOL14 - Thevalingam Donald Homework 14 Due midnight Inst...

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

Thevalingam, Donald – Homework 14 – Due: Dec 21 2006, midnight – Inst: Eslami 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. Please take note oF the due date oF the as- signment. It is due on Thursday, Decembber 21 at midnight. 001 (part 1 oF 1) 10 points An elevator starts From rest with a constant upward acceleration and moves 1 m in the frst 1 . 7 s. A passenger in the elevator is holding a 5 . 5 kg bundle at the end oF a vertical cord. The acceleration oF gravity is 9 . 8 m / s 2 . What is the tension in the cord as the ele- vator accelerates? Correct answer: 57 . 7062 N. Explanation: T mg a elevator g Let h be the distance traveled and a the acceleration oF the elevator. With the initial velocity being zero, we simpliFy the Following expression and solve For acceleration oF the elevator: h = v 0 t + 1 2 at 2 = 1 2 at 2 = a = 2 h t 2 . The equation describing the Forces acting on the bundle is F net = ma = T - mg T = m ( g + a ) = m µ g + 2 h t 2 = (5 . 5 kg) 9 . 8 m / s 2 + 2(1 m) (1 . 7 s) 2 = 57 . 7062 N . keywords: 002 (part 1 oF 3) 10 points A dragster and driver together have mass 881 . 2 kg . The dragster, starting From rest, attains a speed oF 26 . 8 m / s in 0 . 54 s . ±ind the average acceleration oF the drag- ster during this time interval. Correct answer: 49 . 6296 m / s 2 . Explanation: The average acceleration during time t is a = Δ v t = v - 0 t = v t = (26 . 8 m / s) (0 . 54 s) = 49 . 6296 m / s 2 . 003 (part 2 oF 3) 10 points What is the size oF the average Force on the dragster during this time interval? Correct answer: 43733 . 6 N. Explanation: The average Force on the dragster is f d = m d a = (881 . 2 kg)(49 . 6296 m / s 2 ) = 43733 . 6 N . 004 (part 3 oF 3) 10 points Assume: The driver has a mass oF 73 . 2 kg . What horizontal Force does the seat exert on the driver? Correct answer: 3632 . 89 N. Explanation: The Force on the driver is F = ma = (73 . 2 kg)(49 . 6296 m / s 2 ) = 3632 . 89 N .

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

View Full Document
Thevalingam, Donald – Homework 14 – Due: Dec 21 2006, midnight – Inst: Eslami 2 keywords: 005 (part 1 of 2) 10 points The tension in a string from which a 7 . 4 kg object is suspended in an elevator is equal to 40 N. The acceleration of gravity is 9
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

SOL14 - Thevalingam Donald Homework 14 Due midnight Inst...

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

View Full Document
Ask a homework question - tutors are online