A force acts on a resting body weighing 2 kg. After the start of movement, the body has passed 25 meters in 5 seconds

A force acts on a resting body weighing 2 kg. After the start of movement, the body has passed 25 meters in 5 seconds, find the value of the force if the coefficient of friction is 0.02.

We apply the formula:
X: ma = F cos L – Ft;
y: 0 = N + F sin L – mg;
a =
ma = FcosL – M (mg – FsinL);
make transformations:
F =
S = 25;
M = 0.02;
m = 2 kg;
t = 5 s;
Answer: 5 s