Anonymous

# Power required to push lawnmower?

A gardener pushes a 18 kg lawnmower whose handle is tilted up 35° 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 1.4 m/s?

Any thoughts on this one?

Thanks!

Relevance

easy

mass of lawnmower is 18 kilos. so weight is mass times g

w = 18*9.81 = 176.6 N (Newtons)

friction force is normal force, times coefficient, so:

F = w * 0.15 = 26.5 N

work is Force, times Distance, so per second:

w = F * 1.4 = 37.0 J (Joules)

so per second, the gardener has to supply 37 Joules. A Joule per second is also called a Watt.

So the garderner has to supply 37 Watts.

Hope this helps

• Anonymous
4 years ago

Stumped - you and approximately 50 kazillion different boys, my pal. no person makes an adapter for this, that i be attentive to of. some using backyard mowers have a twist/belt combination for turning the belt ninety tiers to a horizontal shaft from a vertical shaft mower. it extremely is likewise executed on some roto-tillers. that must be your only decision. i've got self belief i've got seen this conversion executed on a crude circulate kart, yet regrettably, there are only no PTO kits pre-made to alter the direction of the vertical shaft autos to paintings on circulate-karts that i've got seen in specific made for circulate-kart use. i'm going to guess somebody ought to do extremely nicely making something like this, nonetheless! Horizontal shaft autos are extra costly for some reason, than verticals, it may save some money. possibly. besides, attempt to locate a broken Rototiller or using mower to scavenge for those twist-belt assemblies. stable success! - The Gremlin guy - 3 used circulate-karts, none with autos the two.

• ?
Lv 6

Take on 10 dollar bill, plus the neighbor kid and this equals my fat but on the porch with a lemonade watching the excitement.