Using the diagram above, the effective force pushing it forward at an angle 60º is
-
25.00N
- 28.87N
- 38.50N
- 43.30N
The correct answer is: A
Explanation
The effective force pushing it forward is the horizontal component of the force = F\(_x\) = F Cos\(\theta\)
F\(_x\) = F Cos\(\theta\) = 50 x Cos 60º = 50 x 0.5 = 25N
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