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abdul_shabeer
 3 years ago
A man of weight W is standing on a lift which is moving upward with an acceleration a. The apparent wieght of the man is
abdul_shabeer
 3 years ago
A man of weight W is standing on a lift which is moving upward with an acceleration a. The apparent wieght of the man is

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Aadarsh
 3 years ago
Best ResponseYou've already chosen the best response.2See, I think the new weight would be: W1 = m(g + a) = mg + ma = W + ma = W + (W/g)a Please confirm with other experts also.

abdul_shabeer
 3 years ago
Best ResponseYou've already chosen the best response.0Yes your answer is right

abdul_shabeer
 3 years ago
Best ResponseYou've already chosen the best response.0Can you explain it in much more detail?

abdul_shabeer
 3 years ago
Best ResponseYou've already chosen the best response.0How you got m(g+a)?

Aadarsh
 3 years ago
Best ResponseYou've already chosen the best response.2see, when the object is moving upward, always add the acceleration due to gravity and the new acceleration. Multiply the sum by mass of the object. U will get the force acting on it.

abdul_shabeer
 3 years ago
Best ResponseYou've already chosen the best response.0The force acting on it would be the aapparent wieght of man?

abdul_shabeer
 3 years ago
Best ResponseYou've already chosen the best response.0Will it be the apparent weight of man?

Aadarsh
 3 years ago
Best ResponseYou've already chosen the best response.2I think most probably it should be.

salini
 3 years ago
Best ResponseYou've already chosen the best response.3yes u have given the right formula but how? here it is when the lift is moving up, the normal force is graeter than the weight of man mg but apparent weight or weight in general of a body is determined by the NORMAL FORCE/NORMAL REACTION hence here Nmg=ma N=mg+ma again i repeat when we say the WEIGHT OF A BODY IS SOMETHING IT IS ACTUALLY THE NORMAL REACTION/FORCE AND NOT mg of the body
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