Turner Exam 2 Spring 2005

Turner Exam 2 Spring 2005 - Trevino, Denise – Midterm 2...

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

View Full Document Right Arrow Icon

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

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

Unformatted text preview: Trevino, Denise – Midterm 2 – Due: Mar 9 2005, 10:00 pm – Inst: Turner 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points An elevator has a mass of 1100 kg and car- ries a maximum load of 777 kg . A constant frictional force of 4600 N retards its motion upward. The acceleration of gravity is 9 . 8 m / s 2 . What power must the motor deliver at a instantaneous speed of 3 . 61 m / s if the elevator is designed to provide an upward acceleration of 1 . 16 m / s 2 ? Correct answer: 90870 . 6 W. Explanation: The motor must supply the force ~ T that pulls the elevator upward. Applying Newton’s second law to the elevator gives T- f- M g = M a. Thus T = M ( a + g ) + f = (1877 kg)(1 . 16 m / s 2 + 9 . 8 m / s 2 ) + 4600 N = 25171 . 9 N . and the required instantaneous power is P = T v = (25171 . 9 N)(3 . 61 m / s) = 90870 . 6 W . 002 (part 1 of 1) 10 points A projectile of mass 0 . 596 kg is shot from a cannon, at height 6 . 9 m, as shown in the figure, with an initial velocity v i having a horizontal component of 6 . 3 m / s. The projectile rises to a maximum height of Δ y above the end of the cannon’s barrel and strikes the ground a horizontal distance Δ x past the end of the cannon’s barrel. The acceleration of gravity is 9 . 8 m / s 2 . Δ x v i 5 7 ◦ Δ y 6 . 9m Find the magnitude of the final velocity vector when the projectile hits the ground. Correct answer: 16 . 4025 m / s. Explanation: Let : v x i = 6 . 3 m / s , x i = 0 m , y i = 6 . 9 m , and θ = 57 ◦ . For the horizontal component of vector v i , we have v x i = v i cos θ ; consequently v i = v x i cos θ = (6 . 3 m / s) cos57 ◦ = 11 . 5673 m / s . The work done by gravity depends only on the vertical distance y i , since the work involving Δ y adds and subtracts; i.e. , cancels. The work done by gravity is W = mg y i . From Δ K = 1 2 mv 2 f- 1 2 mv 2 i and the work-energy theorem Δ K = W , the final velocity is v f = r 2 W m + v 2 i = r 2 mg y i m + v 2 i = q 2 g y i + v 2 i Trevino, Denise – Midterm 2 – Due: Mar 9 2005, 10:00 pm – Inst: Turner 2 = h 2(9 . 8 m / s 2 )(6 . 9 m) + (11 . 5673 m / s) 2 i 1 / 2 = 16 . 4025 m / s . Alternate Solution: The initial velocity in the vertical direction is v y i = v x i tan θ = (6 . 3 m / s) tan57 ◦ = 9 . 70115 m / s , and theverticalvelocity v y top = 0atthetop, there- fore we have v 2 y i = v 2 y top + 2 g Δ y , so Δ y = s v 2 y i 2 g = s (9 . 70115 m / s) 2 2(9 . 8 m / s 2 ) = 4 . 80165 m . The magnitude of the final vertical velocity v y f is | v y f | = p 2 g [ y i + Δ y ] = q 2(9 . 8 m / s 2 )[(6 . 9 m) + (4 . 80165 m)] = 15 . 1444 m / s , so v f = q v 2 y f + v 2 x i = q (15 . 1444 m / s) 2 + (6 . 3 m / s) 2 = 16 . 4025 m / s ....
View Full Document

This note was uploaded on 02/09/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

Page1 / 13

Turner Exam 2 Spring 2005 - Trevino, Denise – Midterm 2...

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

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