m₁ = 2.3 kg
θ₁ = 70°
θ₂ = 17°
g = 9.8 m/s²
->The component of the gravitational force on m₁ that is parallel down the incline is:
F₁ = m₁ × g × sin(θ₁)
F₁ = (2.3
kg) × (9.8 m/s²) × sin(70°) = 21.18 N
->The component of the gravitational force on m₂ that is parallel down the incline is:
F₂ = m₂ × g × sin(θ₂)
F₂ = m₂ × (9.8 m/s²) × sin(70°) = m₂ × (2.86 m/s²)
Then the total mass of the system is:
m = m₁ + m₂
m = (2.3 kg) + m₂
If it is given that m₂ slides down the incline, then F₂ must be bigger than F₁,
and so the net force on the system must be:
F = m₂×(2.86
m/s²) - (21.18 N)
Using Newton's second law, we know that
F = m × a
So if we want the acceleration to be 0.64 m/s², then
m₂×(2.86
m/s²) - (21.18 N) = [(2.3 kg) + m₂] ×
(0.64 m/s²)
m₂×(2.86
m/s²) - (21.18 N) = (1.47 N) + m₂×(0.64
m/s²)
m₂×(2.22
m/s²) = (22.65 N)
m₂ = 10.2
kg
