Aman decides to push a 2.0 kg box from the bottom to the top of a 30-degree incline the coeficient of kinetic friction is 0.2 waht is the amount of force required to keep the box moving up the incline at a constant

Explanation:

m = 2 Kg

α = 30º

μ = 0.2

F = ?

We can apply

∑ Fx' = 0   (as v is constant, a = 0)

then

F - Ff - Wx' = 0    ⇒   F = Ff + Wx'

We get the force of friction as follows

Ff = μ * N = μ * Wy' = μ * (W * Cosα) = μ * m * g * Cosα  

⇒  Ff = 0.2 * 2 Kg * 9.81 m/s² * Cos30º = 3.3983 N

then we get the x'-component of W

Wx' = W * Sin α = m * g * Sin α

⇒  Wx' = 2 Kg * 9.81 m/s² * Sin30º = 9.81 N

finally we obtain

F = Ff + Wx' = 3.3983 N + 9.81 N = 13.2083 N

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg