Answer:
(a) The net force is 80.394 N
The acceleration of the crate is 0.804 m/s²
(b) the final velocity of the crate is 5.02 m/s
Explanation:
Given;
mass of the crate, m = 100 kg
applied force, F = 250 N
angle of inclination, θ = 45°
coefficient of friction, μ = 0.12
Applied force in y-direction, ![F_y = Fsin \theta = 250sin45 = 176.78 \ N](https://tex.z-dn.net/?f=F_y%20%3D%20Fsin%20%5Ctheta%20%3D%20250sin45%20%3D%20176.78%20%5C%20N)
Applied force in x-direction, ![F_x = Fcos \theta = 250cos45 = 176.78 \ N](https://tex.z-dn.net/?f=F_x%20%3D%20Fcos%20%5Ctheta%20%3D%20250cos45%20%3D%20176.78%20%5C%20N)
The normal force is calculated as;
N + Fy -W = 0
N = W - Fy
N = (100 x 9.8) - 176.78
N = 980 - 176.78 = 803.22 N
The frictional force is given by;
Fk = μN
Fk = 0.12 x 803.22
Fk = 96.386 N
(a) The net force is given by;
![F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20F_x%20-%20F_k%5C%5C%5C%5CF_%7Bnet%7D%20%3D%20176.78-96.386%5C%5C%5C%5CF_%7Bnet%7D%20%3D%2080.394%20%5C%20N)
Apply Newton's second law of motion;
F = ma
![a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BF_%7Bnet%7D%7D%7Bm%7D%5C%5C%5C%5C%20a%20%3D%20%5Cfrac%7B80.394%7D%7B100%7D%5C%5C%5C%5C%20a%20%3D%200.804%20%5C%20m%2Fs%5E2)
(b) the velocity of the crate after 5.0 s
![F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s](https://tex.z-dn.net/?f=F%20%3D%20ma%3D%20%5Cfrac%7Bm%28v-u%29%7D%7Bt%7D%20%5C%5C%5C%5CFt%20%3Dm%28v-u%29%5C%5C%5C%5Cv-u%20%3D%20%5Cfrac%7BFt%7D%7Bm%7D%5C%5C%5C%5C%20v%20%3D%20%5Cfrac%7BFt%7D%7Bm%7D%20%2B%20u%5C%5C%5C%5Cv%20%3D%20%5Cfrac%7BF_%7Bnet%7D%2At%7D%7Bm%7D%20%2B%20u%5C%5C%5C%5Cv%20%3D%20%5Cfrac%7B80.394%2A5%7D%7B100%7D%20%2B%201%5C%5C%5C%5Cv%20%3D%205.02%20%5C%20m%2Fs)