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
The acceleration of the box is 0.67 m/s² or 2/3 m/s²
The distance is 3 m
here we used the following equation
F-F' = ma............ Equation 1
here
F = Horizontal force applied to the box,
F' = Frictional force,
m = mass of the box,
a = acceleration of the box.
Now we can write as
a = (F-F')/m............ Equation 2
Now
a = (230-210)/30
a = 20/30
a = 0.67 m/s² or 2/3 m/s²
Now the below equation should be used to determine the distance
s = ut+1/2at²............ Equation 3
here
u = initial velocity,
t = time,
a = acceleration,
s = distance.
So,
s = 0(3)+1/2(2/3)(3²)
s = 3 m.
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