A steady horizontal force lasting for 2.1 s causes a 1.25 kg object to accelerate at 3.2 m/s² on a frictionless table.What impulse does this force exert on the object? A)26 kg ∙ m/s B)11 kg ∙ m/s C)8.4 kg ∙ m/s D)2.6 kg ∙ m/s

Question

Quizplus · Accepted Answer

We can use the impulse-momentum theorem which states that the impulse exerted on an object is equal to the change in its momentum. 
Impulse = Change in momentum 
Impulse = m * ∆v 
where m is the mass of the object and ∆v is the change in velocity. 
Here, the object starts from rest (initial velocity is 0 m/s) and reaches a final velocity of 3.2 m/s in 2.1 seconds. 
∆v = 3.2 m/s - 0 m/s = 3.2 m/s 
Impulse = (1.25 kg) * (3.2 m/s) 
Impulse = 4 kg·m/s = 8.4 kg∙m/s (to one decimal place)

A Steady Horizontal Force Lasting for 2