What average force is required to stop a 920kg car in 7.7s if the car is traveling at 83km/h ?
force*time=change in momentum=mv
change km/hr to m/s first.
To find the average force required to stop a car, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
First, let's convert the car's speed from km/h to m/s:
Given:
Mass (m) = 920 kg
Time taken (t) = 7.7 s
Speed (v) = 83 km/h
To convert km/h to m/s, divide by 3.6:
Speed (v) = 83 km/h ÷ 3.6 = 23.056 m/s
Next, we need to find the acceleration (a) using the formula:
a = (final velocity - initial velocity) ÷ time taken
Since we want to stop the car, the final velocity (vf) would be 0:
a = (0 - 23.056 m/s) ÷ 7.7 s
a = -23.056 m/s ÷ 7.7 s
a ≈ -2.999 m/s²
Now that we have the acceleration, we can calculate the force using Newton's second law:
F = m × a
F = 920 kg × (-2.999 m/s²)
F ≈ -2759.08 N
However, it's important to note that the force is a vector quantity, so the negative sign indicates that the force is in the opposite direction of the car's motion.
Therefore, the average force required to stop the car is approximately 2759.08 Newtons, in the direction opposite to the car's motion.