A2.0-kilogram laboratory cart is sliding across a horizontal frictionless surface at a constant velocity of 4.0 meters per second east. what will be the cart's velocity after a 6.0-newton westward force acts on it for 2.0 seconds and in what direction?

v = 2 m/s (West)

Explanation:

Given

m = 2 Kg

v initial = 4 m/s (East)

F = 6 N (West)

t = 2 s

We can use the formula

v = v initial + a*t

if F = m*a  ⇒   a = F/m = 6 N / 2 Kg

a = 3 m/s² (West)

then

v = v initial + a*t = (4 m/s) + (-3 m/s²)(2 s)

v = - 2 m/s = 2 m/s (West)

4

explanation:

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