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
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
2 west
To determine the cart's velocity after a force is applied, we can use Newton's second law of motion. The formula is:
F = m * a
Where:
F is the force applied to the cart
m is the mass of the cart
a is the acceleration of the cart
Since the cart is initially moving at a constant velocity, its acceleration is zero. Therefore, the force applied will not change the cart's velocity.
So, the cart's velocity will remain constant at 4.0 meters per second east, which corresponds to answer choice (1).