Question

A gardener pushes a 20 kg lawnmower whose handle is tilted up 37∘ above horizontal. The lawnmower's coefficient of rolling friction is 0.18. How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.5 m/s ? Assume his push is parallel to the handle.

Answer 1

Answer:

58.4 W

Explanation:

The speed of the lawnmower is constant: this means that its acceleration is zero, so the net force on it is zero.

The equation of the forces along the two directions therefore are:

- Perpendicular to the floor: $F sin \theta + R -mg =0$

- Parallel to the floor: $F cos \theta - \mu R = 0$

where

F is the push of the gardener

R is the normal reaction

m = 20 kg is the mass

g = 9.8 m/s^2 is the acceleration of gravity

$\mu=0.18$ is the coefficient of friction

$\theta=37^{\circ}$

Solving for R,

$R=mg-Fsin \theta$

Substituting into the other equation,

$F cos \theta - \mu (mg-Fsin \theta) = 0\\F cos \theta - \mu mg + \mu F sin \theta = 0\\F(cos \theta+\mu sin \theta)=\mu mg\\F=\frac{\mu mg}{cos \theta + \mu sin \theta}=\frac{(0.18)(20)(9.8)}{cos 37^{\circ}+0.18(sin 37^{\circ})}=38.9 N$

And the power he must supply therefore is the product of this force and the speed:

$P=Fv=(38.9)(1.5)=58.4 W$