Answer:
0.67 m/s² or 2/3 m/s²
3 m
Explanation:
Using
F-F' = ma............ Equation 1
Where F = Horizontal force applied to the box, F' = Frictional force, m = mass of the box, a = acceleration of the box.
make a the subject of the equation
a = (F-F')/m............ Equation 2
Given: F = 230 N, F' = 210 N, m = 30 kg.
substitute into equation 2
a = (230-210)/30
a = 20/30
a = 0.67 m/s² or 2/3 m/s²
The acceleration of the box = 2/3 m/s²
Using,
s = ut+1/2at²............ Equation 3
Where u = initial velocity, t = time, a = acceleration, s = distance.
Given: u = 0 m/s (from rest), t = 3 s, a = 2/3 m/s²
substitute into equation 3
s = 0(3)+1/2(2/3)(3²)
s = 3 m.
Hence the box moves 3 m
