Answer:
A) v = √[(F₀/m - μ₀) 2x + x2 / L]
, b) v = √[2 (F₀ / m - μ₀) +1] L
Explanation:
For this exercise we will use Newton's second law
X axis
Fo - fr = ma
Y Axis
N - W = 0
N = mg
The equation for the force of friction is
fr = μ N
.fr = μ mg
Let's replace
Fo - μ mg = m a
a = Fo / m - μ g
They ask us the speed of the block, so we must use
a = dv / dt = dv/dx dx dt = dv/dx v
v dv = a dx
We replace
v dv = (F₀ / m - μ₀ (1- x/L)) dx
We integrate
∫v dv = F₀ / m ∫ dx - μ₀ ∫ dx + 1 / L ∫ x dx
½ v² = (F₀ / m - μ₀) x + 1 / L x² / 2
We evaluate from the lower limit, for x = 0 the speed is zero v = 0, to the upper limit x, v
½ v² = (F₀ / m - μ₀) x + 1 / 2L x²
v = √[(F₀ / m - μ₀) 2x + x2 / L]
Part B
The velocity for x = L
v = √(F₀ / m - μ₀) 2L + L2 / L
v = √[2 (F₀ / m - μ₀) +1] L
