Answer:
j
Explanation:
x = 4 t^2 - 2 t - 4.5
Position at t = 3 s
x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m
Velocity at t = 3 s
v = dx / dt = 8 t - 2
v ( t = 3 s) = 8 x 3 - 2 = 22 m/s
Acceleration at t = 3 s
a = dv / dt = 8
a ( t = 3 s ) = 8 m/s^2
When is the velocity = 0
v = 0
8 t - 2 = 0
t = 0.25 second
When is the position = 0
x = 0
4 t^2 - 2 t - 4.5 = 0
t = 1.4 second
Answer:
The magnitude of the friction force is also F.
Explanation:
By the second Newton's law, we know that:
F = m*a
Net force is equal to mass times acceleration.
Here, we know that the box moves with constant speed, thus, the box has no acceleration, then the net force applied to the box is zero.
Also, remember that the friction force is given by:
Where mu is the coefficient of friction, and this force opposes to the direction of motion (that coincides with the direction of our forward force, that is why this has a negative sign)
The net force will be equal to the sum of our two horizontal forces (as the weight is already canceled by the normal force)
And this is equal to zero, because we know that the box is non-accelerated.
Then we must have that:
Then we can conclude that the magnitude of the friction force is F.