A 40-kg box is being pushed along a horizontal smooth surface.The pushing force is 15 N directed at an angle of 15° below the horizontal.What is the magnitude of the acceleration of the crate? A)0.16 m/s² B)0.36 m/s² C)0.47 m/s² D)0.39 m/s² E)0.68 m/s²

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Quizplus · Accepted Answer

The horizontal component of the force is $ F \cos(\theta) = 15 \cos(15^\circ) $. The acceleration $ a $ is given by $ a = \frac{F \cos(\theta)}{m} $. Substituting the values, $ a = \frac{15 \cos(15^\circ)}{40} \approx 0.36 \, \text{m/s}^2 $.

A 40-Kg Box Is Being Pushed Along a Horizontal Smooth