A 1532.9 kg car traveling with a speed of 12.7 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 50.4 m?
a=v²/2s
F=ma= mv²/2s
a vertical spring stretches 9.6cm when a 13 kg block is hung from its end. the block is then displaced (a)constant(b)frequency(c)
To solve this problem, we can use the equations of motion.
The first equation we can use is the equation of motion for distance:
s = ut + 0.5at^2
Where:
s = distance traveled (50.4 m)
u = initial velocity (12.7 m/s)
t = time taken to stop (unknown)
a = acceleration (unknown)
Since the car comes to a halt, its final velocity is 0 m/s. Therefore, we can use the equation for final velocity to solve for time:
v = u + at
Substituting the values, we have:
0 = 12.7 + a * t
Now, we have two equations with two unknowns. We can solve for time in the second equation and substitute it into the first equation.
From the second equation:
t = -12.7 / a
Substituting this value of t into the first equation:
50.4 = 12.7 * (-12.7 / a) + 0.5 * a * (-12.7 / a)^2
Simplifying the equation, we get:
50.4 = -12.7^2 / a + 0.5 * a * (12.7^2 / a^2)
Simplifying further, we get:
50.4 = -12.7^2 / a + 0.5 * 12.7^2 / a
To solve this equation, we can multiply both sides by a to eliminate fractions:
50.4a = -12.7^2 + 0.5 * 12.7^2
Now, rearranging the equation:
50.4a = -12.7^2 * (1 - 0.5)
Calculating:
50.4a = -161.29
Finally, we solve for a:
a = -161.29 / 50.4
a ≈ -3.199 m/s^2
Now that we have the acceleration, we can calculate the net force required to bring the car to a halt using Newton's second law:
F = m * a
Substituting the values:
F = 1532.9 kg * (-3.199 m/s^2)
Calculating the force, we get:
F ≈ -4903.55 N
The magnitude of the horizontal net force required to bring the car to a halt is approximately 4903.55 N.