Question
A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 10.0 kg, travels 6.50 m up the conveyor belt at a constant speed without slipping. If the conveyor belt is inclined at a 60.0° angle, calculate the work done on the bag by: The force of gravity (g), the normal force (N), the friction force (f), the conveyor belt, and the net force.
Answers
Wt. = m*g = 10kg * 9.8N/kg = 98 N. = Wt.
of the luggage.
F = 98N.[60o] = Force of luggage.
Fp = 98*sin60 = 84.9 N. = Force parallel to incline.
Fv=98*cos60 = 49 N.=Force perpendicular to incline = Normal.
a. Work=mg * h = 98 * 6.5sin60= 552 J.
b. Work = -98 * 6.5sin60 = -552 J.
c. Work = Fk * d = 0 * 6.5 = 0.
d. Work = Fn*d = 84.9 * 6.5 = 552 J.
of the luggage.
F = 98N.[60o] = Force of luggage.
Fp = 98*sin60 = 84.9 N. = Force parallel to incline.
Fv=98*cos60 = 49 N.=Force perpendicular to incline = Normal.
a. Work=mg * h = 98 * 6.5sin60= 552 J.
b. Work = -98 * 6.5sin60 = -552 J.
c. Work = Fk * d = 0 * 6.5 = 0.
d. Work = Fn*d = 84.9 * 6.5 = 552 J.
sloppy work henry
work by conveyer belt = work of normal + work of friction force why i don't know its dumb.