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A 10 kg block is pulled on a rough horizontal surface ($\mu_k = 0.2$) by a force of 50 N acting at an angle of 37° above the horizontal. What is its acceleration? (Take $g = 10	ext{ m/s}^2, \sin 37^\circ = 0.6, \cos 37^\circ = 0.8$) A). $2	ext{ m/s}^2$ B). $3	ext{ m/s}^2$ C). $2.6	ext{ m/s}^2$ D). $5	ext{ m/s}^2$

Edvaya · Accepted Answer

The correct answer is $2.6	ext{ m/s}^2$. **Solution:** 1. Resolve the pulling force ($F = 50	ext{ N}$) into components: Horizontal: $F_x = 50\cos 37^\circ = 50(0.8) = 40	ext{ N}$ Vertical: $F_y = 50\sin 37^\circ = 50(0.6) = 30	ext{ N}$ 2. Calculate the Normal reaction ($N$): $N + F_y = mg \implies N = 100 - 30 = 70	ext{ N}$ 3. Calculate kinetic friction ($f_k$): $f_k = \mu_k N = 0.2 	imes 70 = 14	ext{ N}$ 4. Apply Newton's Second Law horizontally: $F_{	ext{net}} = F_x - f_k = 40 - 14 = 26	ext{ N}$ $$a = \frac{F_{	ext{net}}}{m} = \frac{26}{10} = 2.6	ext{ m/s}^2$$

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