TRANSCRIBE THE FOLLOWING SOLUTION IN DIGITAL FORMAT PLEASE 0.5+6010.5= 2.857 ft/secforce balonceon the block +bullet(m+mb) d²cdt2=-1²xlet the solution be xlt) = A sinut(m+mb) (-Aw² Sinot) = -K A sinutw=√ K = √ 100+32 +121l bf/ftm+mb10.5lbW= 60.475²1x= A sin (60.474) ftv(t) = dx (t)dtThe= 60.47 A cos (60.47 +) ft/secv (0) = -√ = -2.85760.47xA==2.857A=-0.04725resulting position of block isselt) = -0.04725 Sin (60.477) ftThe resulting Velocity isV(t) = -2.857 Cos (60.477) ft/sec Mass of blockm=1016K=1001b/m=100xglbp/inchmass of bulletvelocity of bulletMb=0-516Vb=60ft/secBy Seeing the diagram WesolutionmmmmKAssumeруmGoft/secg=32 ft/s²Applying conservation ofandxbeforemomentumafter the bullet hits the blockmxmbv= myo +mb+bVv= mbvm+mbfind the

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Answered: TRANSCRIBE THE FOLLOWING SOLUTION IN…

Engineering

Mechanical Engineering

TRANSCRIBE THE FOLLOWING SOLUTION IN DIGITAL FORMAT PLEASE

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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TRANSCRIBE THE FOLLOWING SOLUTION IN DIGITAL FORMAT PLEASE

0.5+60
10.5
= 2.857 ft/sec
force balonce
on the block +bullet
(m+mb) d²c
dt2
=-1²x
let the solution be xlt) = A sinut
(m+mb) (-Aw² Sinot) = -K A sinut
w=√ K = √ 100+32 +121l bf/ft
m+mb
10.5lb
W= 60.475²1
x= A sin (60.474) ft
v(t) = dx (t)
dt
The
= 60.47 A cos (60.47 +) ft/sec
v (0) = -√ = -2.857
60.47xA==2.857
A=-0.04725
resulting position of block is
selt) = -0.04725 Sin (60.477) ft
The resulting Velocity is
V(t) = -2.857 Cos (60.477) ft/sec

Mass of block
m=1016
K=1001b/m
=100xglbp/inch
mass of bullet
velocity of bullet
Mb=0-516
Vb=60ft/sec
By Seeing the diagram We
solution
mmmm
K
Assume
ру
m
Goft/sec
g=32 ft/s²
Applying conservation of
and
x
before
momentum
after the bullet hits the block
mxmbv
= myo +mb+b
V
v= mbv
m+mb
find the

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