JAMB - Physics (1997 - No. 3)
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]
5.0 ms-2
5.8 ms-2
8.7 ms-2
25.0 ms-2
Explanation
ma = mgsin\(\theta\)
a = \(\frac{5 {\times} 10 {\times} sin30}{t}\)
a = 5ms2
a = \(\frac{5 {\times} 10 {\times} sin30}{t}\)
a = 5ms2
Comments (0)
