Ten coins are placed on top of each other on a horizontal table. If the mass of each coin is 10g and acceleration due to gravity is 10 ms^-2, what is the magnitude and direction of the force on the 7^th coin ( counted from the bottom) due to all the coins above it ?

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0.3N downwards

0.3N upwards

0.7N downwards

0.7N upwards

answer is A.

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Detailed Solution

Three coins are above the 7th coin. Therefore, force from 3 coins above it will be

$3 mg = 3 (10 \times 10^{- 3}) (10) = 0.3 N$ (in downward direction)

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