This problem is looking for the minimum value of μs that is
necessary to achieve the record time. To solve this problem:
Assuming the front wheels are off the ground for the entire
¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².
For a constant acceleration, distance = 402.3
m = 1/2at^2 = 804.6 m / (4.43 s)^2 = a = µs·9.8 m/s^2
µs = 804.6 m / (4.43s)^2 / 9.8 m/s^2 = 4.18
Net force
As it's negative the box will move left
<h3>Question:</h3>
How to find g (acceleration due to gravity)
<h3>Solution:</h3>
We know,
Acceleration due to gravity (g)

where, G = Gravitational constant

M = Mass of the earth

R = Radius of the earth

Putting these values of G, M and R in the above formula, we get

So, the value of acceleration due to gravity is

Hope it helps.
Do comment if you have any query.
Answer:
T = 37.08 [N*m]
Explanation:
We must remember that torque is defined as the product of a force by a distance. This distance is measured from the point of application of force to the center of rotation of the rotating body.
The force is equal to the product of mass by gravitational acceleration.
![F=m*g\\F=70*9.81\\F=686.7[N]](https://tex.z-dn.net/?f=F%3Dm%2Ag%5C%5CF%3D70%2A9.81%5C%5CF%3D686.7%5BN%5D)
Now the torque can be calculated:
![T=F*r\\T=686.7*0.054\\T=37.08[N*m]](https://tex.z-dn.net/?f=T%3DF%2Ar%5C%5CT%3D686.7%2A0.054%5C%5CT%3D37.08%5BN%2Am%5D)
V = 1/3 Bh v = 1/3 (13 ac)(43560ft^2/ac)(481ft) v = 90793560 ft^3 * 0.3048m/ft * 0.3048m/ft * 0.3048m/ft = 2570987m^3