Question

How large must be F be in the figure shown to give the 700gm block an acceleration of 30cm/s^2? The coefficient of friction between all surfaces is 0.15

(Q.12)

Answer 1

The force must be 2.18 N large.

Explanation:

Let the tension be T in the string.

Coefficient of friction be 0.15

Let the normal rxn between the block and the ground be N

Let the rxn between the two block be N

F − T − 0.15 (N + N') = 0.7 a  for the larger block.

T− 0.15 N = 0.2 a

N = 0.9 g = 9

N = 0.2 g = 2

T − 0.15 × 0.2 g = 0.2 × 0.3

T = 0.36

F − 0.36 − 0.15 (11) = 0.7 0 x 0.3

F = 2.18 N

Thus the force is 2.18 N

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