need help on number 4 and 5 Chrome File Edit View History Bookmarks People Windo

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need help on number 4 and 5

Chrome File Edit View History Bookmarks People Window Help U.S. 68% 1, Mon Feb 15 6:06 PM a E Homework: Conservative FxTest 1 Sumin WebAssign xMInbox (9) skim526@illinoisxCHome | Chegg.com https://www.flipitphysics.com/Course/ViewProblem?unitltemID-1493828&enrollmentID;=163387 New Tab chem 204 loncapa W math 241 WebAssign P math 225 pearson B chem 205 compass physics 211 flipit Math 241 syllabus IMath 225 syllabus physics 211 syllabus C Chegg ksmrosa@gmail.com| account|log off tech support Apps Monday, February 15 |6:04 PM FlipltPhysics macmillan learning PHYS 211 SPRING 2016 UIUC Physics Unit 8: Prelecture Checkpoint Homework / Problems Homework: Conservative Forces And Potential Energy Print Assignment View Deadline: 100% until Tuesday, February 16 at 8:00 AM Standard Exercise Work Concepts Pendulum 2 (wl 2 Hints) Standard Exercise Pendulum Standard Exercise Pendulum 2 (w/ 2 Hints) Standard Exercise Loop the Loop (w/ Work) Interactive Example IE Block Spring Incline Standard Exercise Exam Practice Question 1 A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass reaches the bottom of its path it is moving at a speed v - 2.3 m/s and the tension in the rope is T 13.7 N 1) How long is the rope? 0.27 Standard Exercise Exam Practice Question 2 Standard Exercise Tipler6 7.P.012 Submit Help Optional

Explanation / Answer

4) a) As v^2= 2gL

length of rope = L =v^2/2g = 2.3^2/2*9.8 = 0.27 m =27 cm

b)   v^2 = 2gh

here h=(4/5)L

v^2 = 2*9.8*0.8*0.27

v = 2.06 m/s

5) a = v^2/(1/5)r =2.06^2/0.2*0.27 = 78.6 m/s^2

After you find acceleration you times it by mass to get the net force of the tension T - 0 = ma --> T = ma

T = ma

Hope that helps

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