The net force on the box parallel to the plane is
∑ F[para] = mg sin(24°) = ma
where mg is the weight of the box, so mg sin(24°) is the magnitude of the component of its weight acting parallel to the surface, and a is the box's acceleration.
Solve for a :
g sin(24°) = a ≈ 3.99 m/s²
The box starts at rest, so after 7.0 s it attains a speed of
(3.99 m/s²) (7.0 s) ≈ 28 m/s
Answer:
22 m/s
Explanation:
v(t) = Vo + at
v(3) = 4 + 6*3
v = 4 + 18
v = 22 m/s (almost 50 miles/hour)
