a 1000 kg car is traveling 27 m/s over 7 seconds what is it average forward force of the car?
27 m/s x 7 s =189 m/s^2 = a
1000 kg x 189 m/s^2 = 189,000 N
v=at 27m/s=a(7s) a= 27/7= 3.86m/s^2 F=ma F= (1000kg)(3.86m/s^2) = 3860 N
FYI I am a physics student rn in college first semester of it. Bear with me if im wrong
That question makes no sense as asked.
Ideally it take ZERO force to maintain a constant speed.
FORCE implies ACCELERATION
F = m A
Now perhaps you mean it accelerated from zero to 27m/s in 7 seconds
a = change in velocity / change in time
= 27/7 =3.86 m/s^2
F = 1000 kg * 3.857 = 3,857 Newtons
Yes CodyJinks :)
To calculate the average forward force of the car, you can use the equation:
Force = mass x acceleration
First, you need to find the acceleration of the car. The formula for acceleration is:
acceleration = (change in velocity) / (time)
In this case, the change in velocity is given as 27 m/s and the time is 7 seconds. So, the acceleration is:
acceleration = 27 m/s / 7 s = 3.86 m/s^2
Next, you can use the equation Force = mass x acceleration. The mass of the car is given as 1000 kg, and the acceleration calculated above is 3.86 m/s^2. So, the average forward force of the car is:
Force = 1000 kg x 3.86 m/s^2 = 3860 N
Therefore, the average forward force exerted by the car is 3860 Newtons.