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

surface. What is the magnitude of the normal force exerted by the surface on the crate?

Answer 1

N = 52.0 N

Step-by-step 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}`