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Question: A block A is made to move over an inclined plane of inclination q with constant acceleration a<sub>0...

A block A is made to move over an inclined plane of inclination q with constant acceleration a0 as shown in figure. Initially bob B hanging from block A is held vertical. Magnitude of acceleration of block A relative to bob immediately after bob is released is –

A

a0

B

a0 sinq

C

a0 cosq

D

(a0 – g sinq)

Answer

a0 cosq

Explanation

Solution

aAG\overrightarrow { \mathrm { a } } _ { \mathrm { AG } } = a0 cosθi^\cos \theta \hat { i } + a0 sinq j^\hat { \mathrm { j } }

aBG=a0sinθj^\overrightarrow { \mathrm { a } } _ { \mathrm { BG } } = \mathrm { a } _ { 0 } \sin \theta \hat { \mathrm { j } }

aAB=a0cosθi^\therefore \overrightarrow { \mathrm { a } } _ { \mathrm { AB } } = \mathrm { a } _ { 0 } \cos \theta \hat { \mathrm { i } }