Work of Lawn mower

$W =F.d. \cos μ$ $Joules$

$W= 80 ×30×\cos30°$ $Joules$

$\boxed{2078.46}$ $Joules$

We're looking for the x component of the work done by the person onto the lawnmower. The formula for work is $W=Fd$ . In this problem, we'll be looking for the x component of the work, so we'll be using the formula $\large{W_x=F_x \cdot x}$ , where $x$ is the horizontal distance over which the person pushes the lawnmower forward $(x=30 \text{ m})$ .

In the image below, the red rectangle is the lawnmower. The force applied by the person is $\large{F_{applied}}$ . $\large{F_x}$ is the x component of the force he applies on the lawnmower, and $\large{F_y}$ is the y component of the person's force (neglecting the force of gravity, since we only care about the force applied by the person).

The image below shows the equations for the $x$ and $y$ components of the applied force. We're only concerned with the x component, so we'll need to solve the blue equation for the x component of the force the person applies onto the lawnmower.

$F_x = 80 \cos{30 ^\circ} = 80 \dfrac{\sqrt 3}{2} = 40 \sqrt 3 \text{ N}$

Now that we know $F_x$ and $x$ , we can solve for $W_x$ :

$\large{W_x= (40 \sqrt 3 \text{ N}) (30 \text{ m}) = 1200 \sqrt 3 \text{ J} \approx \boxed{2078.46 \text{ J}}}$ .