Asked by kayla
A person pushes a 17.0 kg lawn mower at constant speed with a force of 74.0 N directed along the handle, which is at an angle of è = 48.0° to the horizontal
(b) Calculate the horizontal retarding force on the mower
(c) Calculate the normal force exerted vertically upward on the mower by the ground.
(d) Calculate the force the person must exert on the lawn mower to accelerate it from rest to 1.3 m/s in 2.0 seconds (assuming the same retarding force).
(b) Calculate the horizontal retarding force on the mower
(c) Calculate the normal force exerted vertically upward on the mower by the ground.
(d) Calculate the force the person must exert on the lawn mower to accelerate it from rest to 1.3 m/s in 2.0 seconds (assuming the same retarding force).
Answers
Answered by
Henry
Wm = mg = 17kg * 9.8N/kg = 166.6 N. =
Wt. of mower.
Fm = 166.6 N @ 0 Deg. = Force of mower.
Fp = 166.6*sin(0) = 0. = Force parallel
to gnd.
Fv = 166.6*cos(0) = 166.6 N. = Force
perpendicular to Gnd. = Normal.
b. Fn = Fap-Fp-Fr = 0,
74*cos48-0-Fr = 0,
49.52-Fr = 0,
Fr = 49.52 N. = Retarding force.
c. Fv = mg = 166.6 N.
d. a = (Vf-Vo)/t,
a = (1.3-0) / 2 = 0.65 m/s^2.
Fcos48-Fr = ma,
Fcos48-49.52 = 17*0.65 = 11.05,
Fcos48 = 11.05+49.52 = 60.57,
F = 60.57 / cos48 = 90.5 N.
Wt. of mower.
Fm = 166.6 N @ 0 Deg. = Force of mower.
Fp = 166.6*sin(0) = 0. = Force parallel
to gnd.
Fv = 166.6*cos(0) = 166.6 N. = Force
perpendicular to Gnd. = Normal.
b. Fn = Fap-Fp-Fr = 0,
74*cos48-0-Fr = 0,
49.52-Fr = 0,
Fr = 49.52 N. = Retarding force.
c. Fv = mg = 166.6 N.
d. a = (Vf-Vo)/t,
a = (1.3-0) / 2 = 0.65 m/s^2.
Fcos48-Fr = ma,
Fcos48-49.52 = 17*0.65 = 11.05,
Fcos48 = 11.05+49.52 = 60.57,
F = 60.57 / cos48 = 90.5 N.
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