A mass M slides upward along a rough plane surface inclined at angle \( \Theta=0.25 \) in radians to the horizontal. Initially the mass has a speed \( V_{0}=2.96 \mathrm{~m} / \mathrm{s} \), before it slides a distance \( \mathrm{L}=1.0 \mathrm{~m} \) up the incline. After sliding this distance the new speed of the mass is \( V_{0} / 4 \) measured in \( \mathrm{m} / \mathrm{s} \). What is the acceleration of the sliding mass? (Positive denotes acceleration up the incline; negative denotes acceleration down the incline.)
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96 \, \mathrm{m/s} \) - Final speed, \( V_f = \frac{V_0}{4} = \frac{2.96}{4} = 0.74 \, \mathrm{m/s} \) - Distance, \( L = 1.0 \, \mathrm{m} \) - Incline angle, \( \Theta = 0.25 \, \text{radians} \) Show more…
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