Answer:
shown in the attachment
Explanation:
The detailed step by step and necessary mathematical application is as shown in the attachment.
D. Teach the public energy conservation
<h2>Answer: at an angle

below the inclined plane.
</h2>
If we draw the <u>Free Body Diagram</u> for this situation (figure attached), taking into account only the gravity force in this case, we will see the weight
of the block, which is directly proportional to the gravity acceleration
:

This force is directed vertically at an angle
below the inclined plane, this means it has an X-component and a Y-component:



Therefore the correct option is c
So vector A has a magnitude of 7 units in the negative Y direction and the Vector B has a magnitude of 14 units in the positive x. Which means that A has a coordinate of A(0,-7) nad B is B(14,0).
-The direction and the magnitude of A+B is -7 down and 14 forward
Answer:
Fp = 26.59[N]
Explanation:
This problem can be solved using the principle of work and energy conservation, i.e. the final kinetic energy of a body will be equal to the sum of the forces that do work on the body plus the initial kinetic energy.
We need to identify the initial data:
d = distance = 41.9[m]
Ff = friction force = 44.5 [N]
m = mass = 16.3 [kg]
v1 = 1.9 [m/s]
v2 = 12.6 [m/s]
The kinetic energy at the beginning can be calculated as follows:
![E_{k1}= \frac{1}{2}*m*v_{1}^2 \\E_{k1}= \frac{1}{2}*16.3*(1.9)_{1}^2\\E_{k1}= 29.42[J]](https://tex.z-dn.net/?f=E_%7Bk1%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%2Am%2Av_%7B1%7D%5E2%20%5C%5CE_%7Bk1%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%2A16.3%2A%281.9%29_%7B1%7D%5E2%5C%5CE_%7Bk1%7D%3D%2029.42%5BJ%5D)
And the final kinetic energy.
![E_{k2}= \frac{1}{2}*m*v_{2}^2 \\E_{k2}= \frac{1}{2}*16.3*(12.6)^2\\E_{k2}= 1294[J]](https://tex.z-dn.net/?f=E_%7Bk2%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%2Am%2Av_%7B2%7D%5E2%20%5C%5CE_%7Bk2%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%2A16.3%2A%2812.6%29%5E2%5C%5CE_%7Bk2%7D%3D%201294%5BJ%5D)
The work is performed by two forces, the friction force and the pushing force, it is important to clarify that these forces are opposite in direction.
The weight of the cart also performs a work in the direction of movement since the plane is tilted down, this component of the weight of the cart must be parallel to the surface of the inclined plane.
![W_{1-2}=-(44.5*41.9)+(16.3*9.81*sin(17.5)*41.9)+(F_{p}*41.9) \\therefore:\\E_{k1}+W_{1-2}=E_{k2}\\29.42+150.16+(F_{p}*41.9)=1294\\F_{p}=1114.42/41.9\\F_{p}=26.59[N]](https://tex.z-dn.net/?f=W_%7B1-2%7D%3D-%2844.5%2A41.9%29%2B%2816.3%2A9.81%2Asin%2817.5%29%2A41.9%29%2B%28F_%7Bp%7D%2A41.9%29%20%5C%5Ctherefore%3A%5C%5CE_%7Bk1%7D%2BW_%7B1-2%7D%3DE_%7Bk2%7D%5C%5C29.42%2B150.16%2B%28F_%7Bp%7D%2A41.9%29%3D1294%5C%5CF_%7Bp%7D%3D1114.42%2F41.9%5C%5CF_%7Bp%7D%3D26.59%5BN%5D)