Asked by James

A 2.0-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?

(1) 2.0 m/s east (3) 10. m/s east
(2) 2.0 m/s west (4) 10. m/s west
Answered by Damon
call east positive x
initial velocity Vo = +4 m/s
initial momentum = m Vo = +8 kg m/s
Force = -6 newtons
time of impulse = 2 sec.
Force times time = change of momentum = (-6)(2) = -12 newton seconds
so change of momentum is -12
final momentum = 8 - 12 = -4 kg m/s
final speed = -4/2 = -2 m/s or 2 m/s west
Answered by Anonymous
2 west
Answered by jhwefjseb
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