A 5kg block is released from rest on a smooth plane inclined at an angle of 30o to the horizontal. What is the acceleration down the plane? [g = 10ms-2]
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Correct Answer: Option A
Explanation:
ma = mgsin\(\theta\)
a = \(\frac{5 {\times} 10 {\times} sin30}{t}\)
a = 5ms2
ma = mgsin\(\theta\)
a = \(\frac{5 {\times} 10 {\times} sin30}{t}\)
a = 5ms2