Fg = m*g = 20kg * 9.8N/kg = 196 N.
Fp = 196*sin40 = 126 N. = Force parallel to incline.
Fn = 196*cos40 = 150 N. = Normal.
h = L*sin40 = 6.5*sin40 = 4.18 m.
Wg = Fg * h = 196 * 4.18 = 819.3 Joules.
Wn = = Fn * h = 150 * 4.18 = 628 N.
Fp-Ff = m*a = m*0 = 0
126-Ff = 0
Ff = 126 N.= Force of friction.
Wf = Ff * L = 126 * 6.5 = 819 Joules.
A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 20.0 kg, travels 6.50 m up the conveyor belt at a constant speed without slipping. If the conveyor belt is inclined at a 40.0° 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.
1 answer
3 answers