A gardener pushes a 12 kg lawnmower whose handle is tilted up 37deg above horizontal. The lawnmower's coefficient of rollingfriction is 0.15.How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.2 m/s? Assume hispush is parallel to the handle.

We are given mass of lawn mover. We are given the angle of push wrt horizontal. We apply 2nd law of motion, along horizontal and vertical. Since the mover is to be moved with constant velocity, the net horizontal force is also zero.Mass is m=12kg Friction coefficient is μ=0.15 The normal force is given as N=mg+Fsin37 Hence friction is f=μNf=0.15mg+Fsin37 Applying balancing of forces along horizontal, Fcos37=fFcos37=0.15mg+Fsin370.8F=0.1512×9.8+0.602F0.8F-0.09F=17.64F=17.640.71F=24.85 N The speed of lawn mover is v=1.2 m/s Hence the power needed is P=FvP=24.85×1.2P=29.83 W Hence the power needed is 29.83 W.

Answered: A gardener pushes a 12 kg lawnmower…

A gardener pushes a 12 kg lawnmower whose handle is tilted up 37deg 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 1.2 m/s? Assume his push is parallel to the handle.