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 groun

Force F = 280 N Angle with respect to the ground = 40 degrees Weight of the Lawnmower = 350 N With constant velocity, Acceleration is 0 Thus, Forward force Ff = F cos theta = 280 cos40 The frictional force opposing it is defined as (u x Force from pressure) + vertical component of the Force, where u indicates the coefficient of friction. Fb = (u x m x g) + (u x 280sin40) With Ff = Fb, we derive: 280 cos40 = u x ((m x g) + 280sin40) u = 280 cos40 / ((350 x 9.81) + 280sin40) results in 214.49 / () = 0.405 Hence, the coefficient of friction u = 0.405.

The coefficient of friction is: µ = 0.405. Explanation: With an acceleration of 0 (constant velocity), all horizontal forces must equal 0. These forces include: The horizontal component of the force (F1) acting FORWARD, calculated as F1 = F*cosθ = 280cos40 N. The friction force F(fr) is given by µ*(total pressure force). Two forces are responsible for the pressure force (Fp). The vertical component of F equals F*sinθ = 280sin40. Thus, the friction force is expressed as F(fr) = µ*(mg + 280sin40). Since F1 = F(fr), we can equate: 280cos40 = µ*(mg + 280sin40). This leads to µ = 280cos40 / (mg + 280sin40). Hence, µ = 0.405.