While the box of mass 3.0kg is moving at the constant velocity of 0.10 m/s , force F increases to 18.0 N and acts for 0.20 s. The frictional force does not change. Calculate the acceleration produced by the resultant force acting on the box.

6 m/s²

Explanation:

Since no net force acts on the box when the velocity is constant, impulse J = Ft = m(v - u) and where F = force acting on box in 0.2 s = 18.0 N, t = time = 0.20 s, u = initial velocity of  box = 0.10 m/s, v = final velocity of box = ? and m = mass of box = 3.0 kg

So, Ft = m(v - u)

Since v is unknown, we make v subject of the formula. So,

v - u = Ft/m

v = Ft/m + u

= 18.0 N × 0.2 s/3.0 kg + 0.10 m/s

= 3.6 Ns/3.0 kg + 0.10 m/s

= 1.2 m/s + 0.10 m/s

= 1.3 m/s

Since v = 1.3 m/s, the acceleration,a of the box produced by the resultant force is

a = (v - u)/t

= (1.3 m/s - 0.10 m/s)/0.2 s

= 1.2 m/s ÷ 0.2 s

= 6 m/s²

So, the acceleration produced by the resultant force is 6 m/s²

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