What is the magnitude of the average force required to stop an 1500 kg car in 8.0 s if the care is traveling at 95 km/h?
Sum of F = ma?
v(final)=v(initial) –at
0= v(initial) –at
a= v(initial)/t =95000/3600•8=3.3 m/s²,
F(fr) =ma=1500•3.3=4948 N
Yes, you are correct! The equation you have mentioned, F = ma, can be used to solve this problem. The magnitude of the average force required to stop an object can be determined by multiplying its mass and the rate at which it decelerates.
Now, let's break down the problem into two steps:
Step 1: Convert the car's initial velocity from km/h to m/s.
To do this, we need to use the conversion factor that 1 km/h is equal to 0.2778 m/s. So, the car's initial velocity is 95 km/h * 0.2778 m/s/km/h = 26.388 m/s.
Step 2: Calculate the acceleration.
We know that acceleration is the change in velocity divided by the time taken. In this case, the car starts at 26.388 m/s and comes to a stop, so the change in velocity is -26.388 m/s (negative because it's decelerating). The time taken is 8.0 s. Thus, the acceleration is (-26.388 m/s) / 8.0 s = -3.2985 m/s².
Finally, we can use the formula F = ma to calculate the magnitude of the average force:
F = (1500 kg) * (-3.2985 m/s²) = -4947.75 N (negative because the force is opposing the motion).
So, the magnitude of the average force required to stop the 1500 kg car in 8.0 s is approximately 4947.75 N.