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ID: 1358750 • Letter: A
Question
Austin - x University of P www.sapingle x , www.saplingle www saplingle x SSOE-Homew © An Advertisem © The Graph Sho x New Tab www.saplinglearning.com/ibiscms/mod/ibis/view.php?id-2203341 aicuiator -a-eriodic i apie Available From: Not Set Due Date: Late submissions Allowed with 2 100 Question 8 of 10 Incorrect 10/15/2015 12 100 Map ddb sapling learning points possibl per day until 10/19/2015 12 97 A block of mass m 6.68 kg is attached to a spring which is resting on a horizontal frictionless table. The block is pushed into the spring, compressing it by 5.00 cm, and is then released from rest. The spring begins to push the block back toward the equilibrium position at x = 0 cm. The graph shows the component of the force (in N) exerted by the spring on the block versus the position of the block (in cm) relative to equilibrium. Use the graph to answer the following questions 4 95 Points Possible: Grade Category: Description: Policies: 10 100 Graded How much work is done by the spring in pushing the block from its initial position at x=-5.00 cm to x = 2.31 cm? 96 My Homework 100 Number 11.5 Fx (N) 0 You can check your answers 8 What is the speed of the block when it reaches2 x= 2.31 cm? You can view solutions when you c give up on any question. 27 Number You have ten attempts per questio 1.85 m/ s -5 4 3 -1 0 23 4 5 x (cm You lose 5% of the points available answer in your question for each ir attempt at that answer. What is the maximum speed of the block? Number eTextbook 2.12 m S Help With This Topic Previous Give Up & View Solution e Check Answer Next Exit Hint Web Help & Video:s Copyright © 2011-2015 Sapling Learning, Inc.-190 about us careerspartners privacy policy erms of use contact us help Ask me anything 9:10 PM 10/14/2015 32% ENGExplanation / Answer
From the graph ,
spring constant = k = slope of the graph = 6/0.05 = 120 N/m
work done for displacement from x = -5 cm to x = 2.31 cm
work done = change in spring potential energy = (0.5) k (xf2 - xi2) = (0.5) (120) (0.052 - 0.02312) = 0.118 J
part B
Using conservation of energy
Total energy when spring compressed at X = - 5 cm = total energy at x = 2.31
(0.5) k x12 = (0.5) m v2 + (0.5) k x22
(120) (0.05)2 = 6.68 v2 + (120) (0.0231)2
v = 0.188 m/s
maximum speed of block is at equilibrium position
using conservation of energy
energy at x = 0.05 m = kinetic energy at x =0
(0.5) m v2 = (0.5) k x2
6.68 v2 = 120 (0.05)2
v = 0.212 m/s
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