A 121.2 N carton is pulled up a frictionless baggage ramp inclined at 27.0 degre

Question

A 121.2 N carton is pulled up a frictionless baggage ramp inclined at 27.0 degrees above the horizontal by a rope exerting a 73.2 N pull parallel to the ramp's surface. The carton travels 6.40 m along the surface of the ramp. The work on the carton by the rope is 468 J and 0 J by normal force.

A. Calculate the work done on the carton by the gravity

B. What is the net work done on the carton?

C. Suppose that the rope is angled at 42.8 degrees 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

Work done by friction force is: W(f) = F(f) x d, where F(f) is the Frictional force and d is the distance moved, and F(f) = µN, where µ is the coefficient of friction Work done by gravity is: W(g) = F(g) x d, where F(g) is the force of gravity, and F(g)=mg, where m is the mass of the object being moved In this case, we're talking about a piano of mass, m = 330 kg, sliding down 3.6m on an incline that's at 28 degrees. Therefore: W(g) = mgdcosØ, where Ø = 28 degrees W(g) = (330)(9.8)(3.6)(0.883) = 10,279.63 joules Here the work done by Friction is: W(f) = µNd = µmgdsinØ, since here N = F(g)sinØ W(f) = (0.4)(330)(9.8)(3.6)(0.469) = 2186.31 joules

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