A box weighing 20 kg was placed on an inclined plane 5 m long and 2 m high. the box moves with

A box weighing 20 kg was placed on an inclined plane 5 m long and 2 m high. the box moves with acceleration if the friction force on its plane is 80 N.

Given:
S = 5m
H = 2m
m = 20kg
Ftr = 80N
Find: a, acceleration
Decision:
Solving this problem, we use Newton’s second law
1) in total, three forces act on the body: gravity, sliding friction, support reactions
2) mg + Ftr + N = ma (in vector form)
3) now projections:
Oh) N-mgsina = 0
N = mgsina
Ox) mgcosa – Ftr = ma
Consider an inclined plane and find the leg of a right-angled triangle that it forms
25 + 4 = x ^ 2
x = root of 29
x = 5.4
Then, sine alpha = 2/5 = 0.4
Cosine alpha = 5.4 / 5 = 1.08 ≈ 1
Substitute the known values:
20 kg * 10 n / kg * 1 – 80N = 20kg * a
20 N – 80N = 20 kg * a
– 60N = 20 kg * a
a = – 3 N / kg
Modulus: a = 3 N / kg
Answer: a = 3 N / kg