Question

1. A 20.0 N force directed 20.0° above the horizontal is applied to a 6.00 kg crate that is traveling on a horizontal

Answer 1

N = 52.0 N

Explanation:

Given: $F_a= 20.0\:\text{N}=\:\text{applied\:force}$

$m=6.00\:\text{kg}$

$N = \text{normal force}$

The net force $F_{net}$ is given by

$F_{net} = N + F_a\sin 20 - mg=0$

Solving for N, we get

$N = mg - F_a\sin 20$

$\:\:\:\:\:\:= (6.00\:\text{kg})(9.8\:\text{m/s}^2) - (20.0\:\text{N}\sin 20)$

$\:\:\:\:\:\:= 52.0\:\text{N}$