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
The correct answer is: A
Explanation
ma = mgsin\(\theta\)a = \(\frac{5 {\times} 10 {\times} sin30}{t}\)
a = 5ms2