A 10 kg lawn mower is pushed along by a force applied at a 20 degree angle to the horizontal. If a the mower accelerates at 3.4 m/s^2, what is this force?

m = mass of lawn mower = 10 kg

a = acceleration along X-direction = 3.4 m/s²

let the applied force be "F"

from the force diagram, we see that only "F Cos20 " is acting along the horizontal direction.

hence F' = net force along x-direction = F Cos20

net force along x-direction is given as

F' = ma

F Cos20 = ma

inserting the values

F Cos20 = 10 x 3.4

F Cos20 = 34

F = 36.2 N

hence the force applied is 36.2 N

Force = mass x acceleration

Mass will be considered in the horizontal direction

So the component of force in horizontal direction is 10 kg Sin (theta)

Putting the value in above equation we get:

F = 10 kg X Sin (20) X 3.4 m/s^2

= 10 Kg x 0.342 x 3.4 m/s^2

= 11.628 Kg-m/s^2