What is the Work Done on a Carton on an Inclined Ramp?

AI Thread Summary
A 128 N carton is pulled up a frictionless ramp inclined at 30 degrees by a rope exerting a 72 N force parallel to the ramp. The work done by the rope was initially calculated incorrectly due to misunderstanding the angle, which is actually 0 degrees since the force is parallel to the ramp. The correct calculation for the work done by the rope is 324 J. The user struggled with calculating the work done by gravity but eventually resolved all issues. The discussion emphasizes the importance of understanding force angles and their impact on work calculations.
SuPaFly43218
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Homework Statement



A 128 N carton is pulled up a frictionless baggage ramp inclined at 30.0 degrees 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 the rope, gravity, and the normal force of the ramp.

Homework Equations



W= force*cos(theta)* displacement

The Attempt at a Solution



Work done on carton by ROPE W=72cos(30)*5.2 Answer comes out to 324 J, which is wrong.

Work done on carton by GRAVITY W=128cos(30)*5.2 Answer comes out to 576 J, which is wrong...

What am I getting incorrect about this?
 
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Well, i figured out the work done on the carton by the rope because the angle of the rope is NOT 30 degrees, it is parallel, thus 0degrees. Still can't figure out the gravity one though..
 
SuPaFly43218 said:

Homework Equations



W= force*cos(theta)* displacement

Did you start with a sketch showing all of the forces acting on the object, along with the ramp and the angle theta?

Where did you get this equation from? What do the variables in it represent? Is the equation relevant here?
 
NVM, figured them all out!
 
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