Power and angle: lawnmower?

Question

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?


Thanks!
Any suggestions?

Answer 1

I think, therefore i am not certain.....

I know that F=ma, or W=mg. Therefore 18kg x really 9.81, but i'll use 10m/seconds squared. W = 180N, so normal force pushing UP on the lawn mover is 180N. If there was only a horizontal force it would take mew or 0.15 x normal fore so, 180N x 0.15. BUT there is a n angle so 35 is feta, 180N is the adjacent side. use trig function (cos) to find hypotenuse. cos35= 180N/ x. when you get x, it equals the applied force to get the mover to keep on moving or sliding as "they" say.

1.4 m/s will be present when force is applied!!!!!!!!!!

Answer 2

For this problem the angle of the lawn mower handle is extraneous. Only the component of force in line with the velocity produces power or work.
P = F(ds/dt)
F = 1.15*18*9.80662 = 26.477874 N
ds/dt = 1.4m/s
P = 1.4*1.15*18*9.80662
P = 38.9 J/s = 38.9 watts

Answer 3

preaty complicated I personaly dont like word problems ask a math teacher goodm luck buddy.