Question

if a block A is moving with acceleration of 5ms^2 , the acceleration of B w.r.t. ground is

a) 5ms^2
B)5√2ms^2
C )5√5ms^2
d)10ms^2
figure is attached..
pls solve ​

Answer 1

$\bf \to \: \green{{ \underline{given \div }}}$

$\sf \to \: block \: a \: is \: moving \: with \: acceleration \: of \: = 5 {ms}^{2}$

$\bf \to \: \orange{{ \underline{to \: find \: the \: acceleration \: of \: b \: w.r.t \: ground}}}$

$\bf \to \: \pink{{ \underline{step \: - by \: - step \: - explanation}}}$

$\sf \to \: acceleration \: of \: a \: = 5 {ms}^{2} \\ \\ \sf \to \: by \: using \: virtual \: work \: method \\ \\ \sf \to \: acceleration \: of \: b \: w.r.t \: a \: is \: = 10 {ms}^{2} downward \\ \\ \sf \to \: b \: has \: an \: acceleration \: = 5 {ms}^{2} in \: horizontal \: direction \: along \: with \: a \\ \\ \sf \to \: so \: net \: acceleration \: of \: b \: is \\ \\ \sf \to \: \sqrt{ {10}^{2} + {5}^{2} } = \sqrt{125} = 5 \sqrt{5} ms {}^{ - 2}$

$\bf \to \: { \underline{method \: = 2}}$

$\sf \to \: when \: you \: assume \: block \: a \: is \: reference \\ \\ \sf \to \: let \: the \: tension \: of \: string \: = t_{1} \: and \: t_{2} \: and \: t_{3} \: and \: t_{4} \: is \\ \\ \sf \to \: let \: the \: string \: is \: = l_{1} \: + l_{2} \: + l_{3} \: + \: l_{4} \\ \\ \sf \to \: by \: using \: the \: double \: differentiate \: we \: get \\ \\ \sf \to \: a_{1} \: + a_{2} \: + a_{3} \: + a_{4} \: = 0 \\ \\ \sf \to \: ( - a) + 0 + ( - a) + ( + b) = 0 \\ \\ \sf \to \: - 2a + b \: = 0 \\ \\ \sf \to \: b \: = 2a \\ \\ \sf \to \: since \: block \: b \: is \: always \: in \: contact \: to \: block \: a \\ \\ \sf \to s_{x} \: = ( v_{net}) _{x} = 0 \\ \\ \sf \to \: a - c = 0 \implies{a \: = c} \\ \\ \sf \to \: \sqrt{ { a}^{2} + {b}^{2} } = \sqrt{ {a}^{2} + {2a}^{2} } \\ \\ \sf \to \: \sqrt{5a {}^{2} } = a \sqrt{5} \\ \\ \sf \to \: a \: = 5ms {}^{2} = given \\ \\ \sf \to \: 5 \sqrt{5}ms {}^{2} \\ \\ \sf \to \: \green{{ \underline{the \: acceleration \: of \: b \: w.r.t \: ground \: = 5 \sqrt{5}ms {}^{2} } }}$

Answer 2

$\small\sf\blue{Given}$

▪︎a block A is moving with accleration of 5ms²

$\small\sf\blue{Answer}$

▪︎option C

$\small\sf\blue{step\:by\:explanation\:}$

▪︎lengthofstring $l1+l2+l3+l4$

▪︎l=$l1+l2+l3+l4$

▪︎$\small\sf\blue{By\: double\: differentiation\: w.r.t t,\:}$

▪︎$\small\sf\blue{we\:have\:}$

▪︎0=$a1+a2+a3+a4$

▪︎⇒$0=a+0+a+a4$

▪︎$−2a=a4$

▪︎$\small\sf\blue{wrt \:ground\:}$

▪︎$\small\sf\blue{acceleration \:of\: Block-B\:}$

▪︎⇒$aB=a4+a$

▪︎⇒$aB=∣a4+a$

▪︎⇒$aB=a42$

▪︎⇒aB=√a42+a2+2a4−a(d)

▪︎⇒aB=√4a2+a2+2a(2a)co(90∘)

▪︎⇒$aB=5a$

▪︎As $a=5m/s2$

▪︎aB=5√5m/s

▪︎Option −C