A 127.6 N carton is pulled up a frictionless baggage ramp inclined at 27.0 above

Question

A 127.6 N carton is pulled up a frictionless baggage ramp inclined at 27.0 above the horizontal by a rope exerting a 76.6 N pull parallel to the ramp's surface. The carton travels 5.60 malong the surface of the ramp.

Calculate the work done on the carton by the rope.

Calculate the work done on the carton by the gravity.

Calculate the work done on the carton by the normal force of the ramp.

What is the net work done on the carton?

Suppose that the rope is angled at 47.6 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?

Explanation / Answer

[note: work done by a force is given as F*d*cos(theta) where F is magnitude of the force, d is displacement and theta is angle between F and d.

so we can resolve the force along the direction of motion so that component along the direction of motion will be F*cos(theta) and work done=component*displacement

similarly for component perpendiuclar to displacement, angle will be 90 degree==>cos(90)=0==>work done=0

hence to find work done by a force, find the component of the force along the displacement and then multipy by displacement to get the work done by the force ]

forces acting on the carton:

component of its weight along the ramp's surface=127.6*sin(27)=57.93 N

component of its weight perpendicular to ramp's surface=127.6*cos(27)=113.69 N

force exerted by the rope=76.6 N, parallel to the ramp's surface and directed upwards

part a:

work done on the carton by the rope=force exerted by the rope*distance travelled

=76.6*5.6=428.96 J

part B:

as component of weight of the carton along the surface acts in opposite direction to the movement,

work done will be negative

work done=force*distance

=-57.93*5.6=-324.41 J

part C:

as normal force is perpendicular to the direction of motion,

work done=force*distance*cos(90)=0

part D:

net work done=428.96-324.41=104.55 J

part E:

assuming pulling force remain unchanged at 76.6 N, direction of pulling force with the ramp=47.6-27=20.6 degrees

then component of pulling force along the surface of the ramp=76.6*cos(20.6)=71.7 N

then work done=71.7*5.6=401.53 J

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