A 30.0-kg box is being pulled across a carpeted floor by a horizontal force of 230 N , against a friction force of 210 N . What is the acceleration of the box? How far would the box move in 3 s , if it starts from rest?

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

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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