6. A 1580 kg car is traveling with a speed of 15.0 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 50.0 m?

Kinetic energy of moving vehicle

=(1/2)mv²
=(1/2)1580*15²
=177750 J
Work done by braking force, F
=F.d
=50F
Equate energies:
50F=177750J
F=177750/50 N
=3555N

To find the magnitude of the horizontal net force required to bring the car to a halt in a distance of 50.0 m, we can use the equations of motion.

First, we need to find the deceleration of the car. We can use the equation:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s, since the car comes to a halt)
u = initial velocity (15.0 m/s)
s = distance (50.0 m)
a = acceleration (unknown, we need to find this)

Rearranging the equation, we have:

a = (v^2 - u^2) / (2s)

Substituting the given values, we get:

a = (0^2 - 15.0^2) / (2 * 50.0)

Simplifying this expression, we get:

a = (-225) / 100

a = -2.25 m/s^2

The negative sign indicates that the car is decelerating, meaning it is slowing down.

Next, we can calculate the force using Newton's second law:

F = ma

Where:
F = force (unknown, we need to find this)
m = mass of the car (1580 kg)
a = acceleration (-2.25 m/s^2)

Substituting the given values, we have:

F = 1580 kg * (-2.25 m/s^2)

Simplifying this expression, we get:

F = -3555 N

Therefore, the magnitude of the horizontal net force required to bring the car to a halt in a distance of 50.0 m is 3555 N.