Block X and Y are attached to each other by a light rope and can slide along a horizontal
surface. Mass of block X is 10 kg and that of block Y is 5 kg. The magnitude of force of friction
on blocks X and Y is 8.0 N and 4.0 N respectively. Find the action-reaction forces that the
blocks exert on each other if an applied force of 40 N [right] acts on block per the illustration
below.

a) The force of X on Y is 7.4 N [right] and the force of Y on X is 4.1 N [left]
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Ob) The force of X on Y is 12 N [right] and the force of Y on X is 14 N [left]
c) The force of X on Y is 14 N [right] and the force of Y on X is 14 N [left]
O d) The force of X on Y is 22 N [right] and the force of Y on X is 22 N [left]

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Respuesta :

onyebuchinnaji onyebuchinnaji · Answer 1 · 2022-07-13T19:57:53+02:00

B. The force of X on Y is is approximately 13 N [right].

The action-reaction forces that the blocks exert on each other is calculated as follows;

Net force on the blocks = 40 N - (8 N + 4 N) = 28 N

Total mass of the blocks = 10 kg + 5 kg = 15 kg

a = F(net)/M

a = 28 N / 15 kg

a = 1.867 m/s²

40 N - 8N - T = 10 x 1.867

T = 13.33 N

40 N - 4N - T = 5 x 1.867

T = 26.67 N

Thus, the force of X on Y is is approximately 13 N [right].

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