A 1.2x10^4-kg truck is travelling south at 22m/s. What net force is required to bring the truck to a stop at 330 m? What is the cause of this net force?
To calculate the net force required to bring the truck to a stop, we can use the equation: net force = mass x acceleration.
First, let's calculate the acceleration of the truck using the initial velocity (22 m/s) and the distance traveled (330 m).
1. We use the formula: distance = (initial velocity^2 - final velocity^2)/(2 x acceleration) to calculate acceleration.
Rearranging the formula, we have:
acceleration = (initial velocity^2 - final velocity^2)/(2 x distance)
2. Plugging in the values, we get:
acceleration = (22^2 - 0)/(2 x 330)
Simplifying further:
acceleration = 484/660
acceleration = 0.733 m/s^2
Now, we can calculate the net force.
3. Using the equation: net force = mass x acceleration, we can substitute the values:
net force = (1.2 x 10^4 kg) x (0.733 m/s^2)
Calculating the multiplication:
net force = 8796 N
Therefore, the net force required to bring the truck to a stop at a distance of 330 m is 8796 N.
The cause of this net force is the friction between the truck's tires and the road, which acts in the opposite direction to the truck's motion and eventually brings it to a stop.
To find the net force required to bring the truck to a stop, we need to calculate the deceleration of the truck using the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s since the truck comes to a stop)
u = initial velocity (22 m/s)
a = acceleration (negative as it is deceleration)
s = distance (330 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Plugging in the values, we get:
a = (0^2 - 22^2) / (2 * 330)
a = (-484) / 660
a ≈ -0.733 m/s^2
Since the acceleration is negative, it indicates deceleration (opposite direction to velocity) in this case.
Now, we can calculate the net force using Newton's second law of motion:
F = ma
Where:
F = net force
m = mass of the truck (1.2x10^4 kg)
a = acceleration (-0.733 m/s^2)
Plugging in the values, we get:
F = (1.2x10^4 kg) * (-0.733 m/s^2)
F ≈ -8,796 N
Therefore, the net force required to bring the truck to a stop is approximately -8,796 N.
The cause of this net force is friction. Friction acts in the opposite direction to the motion of the truck and slows it down. Other factors like air resistance and rolling resistance can also contribute to the net force required to stop the truck, but in this case, we assume friction is the dominant force.
Wouldnt it normally be a braking force?
Vf^2=Vi^2+2ad= Vi^2+2(Force/mass)d
where force and d are in the same direction, it should work out to be negative.