Question

A gardener pushes a 15 kg lawnmower whose handle is tilted up 37∘ above horizontal. The lawnmower's coefficient of rolling friction is 0.21. How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.0 m/s ? Assume his push is parallel to the handle.

Answer 1

From the problem statement: {Mower Mass} = M = (15 kg)
Handle Angle = θ = (37 deg)
Coeff Friction = μ = (0.21)
Speed} = v = (1.0 m/s)
Gardener's Force on Mower= F

Friction Force = F*cos(θ)

Friction Force = μ*(M*g + F*sin(θ))
Equating the 2 above equations:
F*cos(θ) = μ*(M*g + F*sin(θ))
⇒ F*cos(θ) = μ*M*g + μ*F*sin(θ)
⇒ F*(cos(θ) - μ*sin(θ)) = μ*M*g
⇒ F = μ*M*g/(cos(θ) - μ*sin(θ))
⇒ F = M*g/(cos(θ)/μ - sin(θ))
⇒ {Power} = F*v = v*M*g/(cos(θ)/μ - sin(θ))
⇒ {Power} = (1.0 m/s)*(15 kg)*(9.81 m/sec^2)/(cos(37 deg)/(0.21) - sin(37 deg))
⇒ {Power} = 45.92 Watts