iPad 4:55 AM 30%. Block Sliding A block with man m-12 kg rests on a frictionless
ID: 3308910 • Letter: I
Question
iPad 4:55 AM 30%. Block Sliding A block with man m-12 kg rests on a frictionless table and is acceierated by a spring with spring comtant k-4805N after being compresseddstance . 0 409 m from the spring's unstretched length. The noor-5 frictionless encept for arough patch a dstance d . 2 m long. For this nough path, the coefficient of friction . 0.44. Block Siding 2 1)How much work is done by the spring 85 accelerates thhe block 401.89 Your subnn401.89 Computed value:4018 Submitted: Tunday, February 27 at 3:04 AM 2)What is the speed of the block right after it leaves the sring 8.18 Computed value: 8.18 umitted. Tesday, Febnary 27 3 36AM Feedback: Correctt 3) How much work is done by friction as the block crosses the rough spott 103.48 your submissions 403.48! Computed value:-103.45 Submitted Tuesday, February 27 t 4:19 AM 4) What is the speed of the black after passes the rough spot -9.15 Your submalions:915 Computed value: 9.15 Submitted Tuesday, February 27 at 4:40AM 5) Instead, the spring is only compressed a distane-0.118 m before beling released How far into the rough path does the block slide before coming to rest What dihtance does the spring meed to be compresed so that the block will junt barely make it past the rough patch 7)lfthe spring was compressed three tines tarther and then block is released,the work done on the block by the Othe sameExplanation / Answer
4)
Here
W=W1+W2=(1/2)mV2
401.89-103.48=(1/2)*12*V2
V=7.05 m/s
5.
Work done by the spring during the acceleration
W=(1/2)Kx2=umgd
(1/2)*4805*0.1182=0.44*12*9.81*d
d=0.6458 m
6.
(1/2)Kx2=umgd
(1/2)*4805*x2=0.44*12*9.81*2
x=0.2077 m
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