A person pushes on the handle of a lawnmower with a force of 280. n. if the handle makes an angle of 40.0 degrees with the ground, calculate the coefficient of friction if the lawnmower weighs 350. n and is moving at a constant velocity.

Force F = 280 N Angle with the ground = 40 degrees Weight of the Lawnmower = 350 N Velocity is constant so Acceleration is 0 So Forward force Ff = F cos theta = 280 cos40 Frictional force with resists to back Fb = (u x Force from pressure) + vertical component of Force, where u is the coefficient of friction. Fb = (u x m x g) + (u x 280sin40) AS Ff = Fb = > 280 cos40 = u x ((m x g) + 280sin40) u = 280 cos40 / ((350 x 9.81) + 280sin40) = 214.49 / () = 0.405 So the coefficient of friction u = 0.405