A 1000kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to a rest uniformly in a distance of 150 m. What is the magnitude applied to the car to bring it to rest?
vf^2=vi^2+2ad wehre a=force/mass
solve for force.
To find the magnitude of the force applied to bring the car to rest, we can use the concept of deceleration.
Given:
Mass of the car (m) = 1000 kg
Initial velocity (u) = 20.0 m/s
Final velocity (v) = 0 m/s
Distance covered (s) = 150 m
First, we need to find the deceleration (a) of the car using the equation:
v^2 = u^2 + 2as
Rearranging the equation to solve for acceleration, we have:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (0^2 - 20.0^2) / (2 * 150)
Simplifying:
a = (-400) / 300
a = -1.33 m/s^2 (Note: The negative sign indicates deceleration)
Now, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) times acceleration (a):
F = ma
Substituting the values:
F = 1000 kg * (-1.33 m/s^2)
Simplifying:
F = -1330 N
The magnitude of the force applied to bring the car to rest is 1330 N.