cw11 - Karen iF it has an inclination oF 33 . 9 ? Correct...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
patel (kp9583) – CW11 – Mackie – (20211) 1 This print-out should have 3 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. NOTE DEADLINE!! MON 9/12 001 10.0 points A man cleaning his apartment pushes a vac- uum cleaner with a Force oF magnitude 66 . 2 N. The Force makes an angle oF 31 . 7 with the hor- izontal ±oor. The vacuum cleaner is pushed 2 . 44 m to the right along the ±oor. Calculate the work done by the 66 . 2 N Force. Correct answer: 137 . 43 J. Explanation: Using the defnition oF work, we have W = F s cos θ = (66 . 2 N) (2 . 44 m) cos31 . 7 = 137 . 43 J . Note: The normal Force N , the weight mvg , and the upward component (66 . 2 N) sin31 . 7 oF the applied Force do no work because they are perpendicular to the displacement. 002 10.0 points Karen has a mass oF 51 . 4 kg as she rides the up escalator at Woodley Park Station oF the Washington D.C. Metro. Karen rode a distance oF 60 . 1 m, the longest escalator in the Free world. The acceleration oF gravity is 9 . 8 m / s 2 . How much work did the escalator do on
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Karen iF it has an inclination oF 33 . 9 ? Correct answer: 16884 . 9 J. Explanation: d h The escalator raised Karen a height oF h = d sin so the work done on Karen by the escalator is W = mgh = mgd sin = (51 . 4 kg) (9 . 8 m / s 2 ) (60 . 1 m) sin33 . 9 = 16884 . 9 J 003 10.0 points It takes 3 . 87 J oF work to stretch a Hookes-law spring 8 . 5 cm From its unstressed length. How much the extra work is required to stretch it an additional 6 . 03 cm? Correct answer: 7 . 43848 J. Explanation: We begin by determining the value oF k From the energy equation solved For k , k = 2 U x 2 = 2(3 . 87 J) (0 . 085 m) 2 = 1071 . 28 J / m 2 We can now determine the energy at the ad-ditional displacement, U add = 1 2 k x 2 add = 1 2 1071 . 28 J / m 2 (0 . 085 m + 0 . 0603 m) 2 = 11 . 3085 J The extra work required is just the dierence in energy between the two displacements. W = U = U add-U = 11 . 3085 J-3 . 87 J = 7 . 43848 J...
View Full Document

This note was uploaded on 12/10/2009 for the course PHYSICS 2021 taught by Professor Mihalisin during the Fall '09 term at Temple.

Ask a homework question - tutors are online