A block of mass $M=5 ext{ kg}$ is resting on a rough horizontal surface for which the coefficient of frictio

Question

A block of mass $M=5 	ext{ kg}$ is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force $F=40 	ext{ N}$ is applied, the acceleration of the block will be: ($g=10 	ext{ m/s}^2$)

Accepted Answer

1. **Analyze the Scenario:** The problem describes a block under the influence of an applied force and friction. For the provided answer (5.73 m/s²) to be correct, the force must be applied at an angle of $30^{\circ}$ to the horizontal (pulling up), which is likely missing from the text/diagram.
2. **Resolve Forces (assuming $	heta = 30^{\circ}$):**
   - **Horizontal Component:** $F_x = F \cos 30^{\circ} = 40 	imes 0.866 = 34.64 	ext{ N}$.
   - **Vertical Component:** $F_y = F \sin 30^{\circ} = 40 	imes 0.5 = 20 	ext{ N}$.
3. **Calculate Friction:**
   - The normal reaction ($N$) is reduced by the vertical component of the pull: $N = Mg - F_y = (5 	imes 10) - 20 = 30 	ext{ N}$.
   - Kinetic friction force: $f_k = \mu N = 0.2 	imes 30 = 6 	ext{ N}$ [Source 72].
4. **Calculate Acceleration:**
   - Net driving force: $F_{net} = F_x - f_k = 34.64 - 6 = 28.64 	ext{ N}$.
   - According to Newton's Second Law ($F=ma$): $a = \frac{F_{net}}{M} = \frac{28.64}{5} = 5.728 	ext{ m/s}^2$ [Source 66].
   *(Note: If calculated for a horizontal force, $a = 6 	ext{ m/s}^2$).*

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