What net force would actually be necessary to stop an automobile of mass 880 traveling initially at a speed of 52.0 in a distance equal to the diameter of a dime, which is 1.8 ?
You gave no units.
I will assume 880 kg
52 m/s
1.8 cm which is .018 meters
F = m a so we need a
v = Vi + a t
0 = 52 + a t
so a t = -52
a =-52/t
then
d = (1/2) a t^2
.018 = (1/2) (-52/t) t^2
t = .018/26 = 6.92 * 10^-4
so
a = -52/t = -52 /(6.922*10^-4)
then
F = 880 a
This question was asked three days ago, with the proper units included.
See
http://www.jiskha.com/display.cgi?id=1268561532
To find the net force required to stop the automobile, we can use Newton's second law of motion, which states that force (F) equals mass (m) multiplied by acceleration (a), or F = ma.
First, let's find the acceleration of the car using the formula for acceleration:
acceleration (a) = (final velocity - initial velocity) / time
Since we want to stop the car, the final velocity will be 0 m/s. The initial velocity is given as 52.0 m/s. The time taken to stop the car is not given, but we can assume that it happens very quickly, almost instantaneously (0 seconds).
a = (0 - 52.0) / 0
a = -52.0 / 0
This results in an undefined value since we cannot divide by zero.
However, we can still find the net force required to stop the car by assuming a small finite time, such as 0.01 seconds. This will give us an approximation of the force needed to stop the car.
a = (0 - 52.0) / 0.01
a = -5200 m/s^2
Now that we have the acceleration, we can calculate the net force using Newton's second law:
F = ma
F = (880 kg) * (-5200 m/s^2)
F = -4,576,000 N
Therefore, the net force required to stop the automobile of mass 880 kg, initially traveling at a speed of 52.0 m/s, in a distance equal to the diameter of a dime, which is 1.8 cm, would be approximately -4,576,000 Newtons.