A gardener pushes a $12kg$ lawnmower whose handle is tilted up $37°$ above horizontal. The lawnmower’s coefficient of rolling friction is $0.15$. How much power does the gardener have to supply to push the lawnmower at a constant speed of $1.2m/s$? Assume his push is parallel to the handle.

According to Newton's second law the net force acting on object is equal to its product of mass and acceleration. Therefore, the net force acting on the object is

$\sum _{}F=ma$

The weight of the force acting on the object due to gravity is given as

$F_g=mg$

The friction force acting on the object is

$f_k=\mu N$

The power required to push the object with the force F is

$P=Fv\cos \left(\theta \right)$

Draw the free body diagram of lawnmower.

As the lawnmower is moving with constant speed, the acceleration of the lawnmower is zero which also says that net force acting on it is zero.

From the above figure, the force is acting in vertical direction $N=mg+F\sin \left(\theta \right)$

Substitute $mg+F\sin \left(\theta \right)$for N in the equation $f_k=\mu N$

Therefore, $f_k=\mu (mg+F\sin \left(\theta \right))$

Refer to above figure and balance the force acting in horizontal direction,

$F\cos \left(\theta \right)=f_k$

Rearrange the equation for F,

$F=\frac{\mu mg}{\cos \left(\theta \right)-\mu \sin \left(\theta \right)}$

The power supplied by the gardener to push the lawnmower at a constant speed v can be determined by using following equation.

$$\begin{gathered} P=Fv\cos \left(\theta \right) \\ P=\left(\frac{\mu mg}{\cos \left(\theta \right)-\mu \sin \left(\theta \right)}\right)v\cos \left(\theta \right) \\ P=\left(\frac{(0.15)(12kg)(9.81m/s^2)}{\cos \left(37°\right)-(0.15)(\sin \left(37°\right)}\right)(1.2m/s)37° \\ P=23.87W \\ P=24W \end{gathered}$$