A gardener pushes a 15 k g lawnmower whose handle is tilted up 37 ? above horizo

Question

A gardener pushes a 15kg lawnmower whose handle is tilted up 37? above horizontal. The lawnmower's coefficient of rolling friction is0.18. How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.5m/s ? Assume his push is parallel to the handle.
A gardener pushes a 15kg lawnmower whose handle is tilted up 37? above horizontal. The lawnmower's coefficient of rolling friction is0.18. How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.5m/s ? Assume his push is parallel to the handle.
A gardener pushes a 15kg lawnmower whose handle is tilted up 37? above horizontal. The lawnmower's coefficient of rolling friction is0.18. How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.5m/s ? Assume his push is parallel to the handle.

Explanation / Answer

normal reaction N = mg + Fsin(37)

f = .18* N

.18N = Fcos(37)

.18(mg+Fsin(37)) = Fcos(37)

F = 38.37 N

Power = Fcos(37)*V = 38.37cos(37)*1.5 = 45.965 W

Update

as I said in comment

Friction force must balance horizontal component of Push force to move at constant speed

friction force = uN = u(mg+Fsin(37))

horizontal component = Fcos(37)

u = coefficient of friction = .18

m = 15

g = 9.81

solver for F

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