Answer:
Explanation:
For this problem we use the translational equilibrium condition. Our reference frame for block 1 is one axis parallel to the plane and the other perpendicular to the plane.
X axis
-Aₓ - f_e +T = 0 (1)
Y axis
N₁ - W_y = 0 ( 2)
let's use trigonometry for the weight components
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
We write the diagram for the second body.
Note that in the block the positive direction rd upwards, therefore for block 2 the positive direction must be downwards
W₂ -T = 0 (3)
we add the equations is 1 and 3
- W₁ sin θ - μ N₁ + W₂ = 0
from equation 2
N₁ = W₁ cos θ
we substitute
-W₁ sin θ - μ (W₁ cos θ) + W₂ = 0
W₂ = m₁ g (without ea - very expensive)
This is the smallest value that supports the equilibrium system
Do you have a picture of the diagram that I could view?
For Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:

Rewriting the acceleration as the increment of velocity

in a time

:

, F becomes

But given the definition of momentum:

, then

represents the momentum change. So we can rewrite F as

And re-arranging the formula we can calculate the value of the change in momentum:
The car would go from zero to 58.0 mph in 2.6 sec.
Since the force on the car is constant, therefore the acceleration of the car would also be constant.
Now for constant acceleration we can use the equation of motion
Using first equation of motion to calculate the acceleration of the car
v=u+at
29=0+a×1.30 ...... Eq. (1)
Again using the first equation of motion
58=0+a*t ....... Eq. (2)
Dividing eq. (2) with equation 1
t=2×1.3
t=2.6 sec