mass of particle is 0.5kg ,it is moving initially with a speed of 80m/s towards east.at t=0 , when particle is at x=0,a force of 20N directed towards west is being applied on it for 4sec , it's position after 5sec is?
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answer -80
Explanation:
F = ma so acceleration comes to 40 m/s^2 now as finally the block has to stop so v= 0 and using third equation of motion s = v^2 - u^2 / 2a i.e ( 80)^2 / 2 × 40 = -80
thus the position of block is -80
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