A128.0-n carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-n pull parallel to the ramp’s surface. if the carton travels 5.20 m along the surface of the ramp, calculate the work done on it by (a) the rope, (b) gravity, (c) the normal force of the ramp. (d) what is the net work done on the carton? (e) suppose that the rope is angled at 50.0∘above the horizontal, instead of being parallel to the ramp’s surface. how much work does the rope do on the carton in this case?

(A) 374.4 J

(B) -332.8 J

(C) 0 J

(D) 41.6 J

(E)  351.8 J

Explanation:

weight of carton (w) = 128 N

angle of inclination (θ) = 30 degrees

force (f) = 72 N

distance (s) = 5.2 m

(A) calculate the work done by the rope

        the ramp θ will be 0

       work done = 72 x 5.2 x cos 0

       work done by the rope = 374.4 J

(B) calculate the work done by gravity

      work done by gravity = 128 x 5.2  x cos 120

      work done by gravity = -332.8 J

(C) find the work done by the normal force acting on the ramp

         work done by the normal force = Fn x distance x cos θ

         work done by the normal force = Fn x 5.2 x cos 90

         work done by the normal force = Fn x 5.2 x 0

         work done by the normal force = 0 J

(D)  what is the net work done ?

     net work done = 374.4 - 332.8 + 0 =  41.6 J  

(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal

work done = 72 x 5.2 x cos 20

work done = 351.8 J

would be false hope this out

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr