Solving Work and Energy Questions: Skier of Mass 67kg

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SUMMARY

The discussion focuses on calculating the work required to pull a skier of mass 67 kg up a 30° slope over a distance of 40 m at a constant speed of 1.7 m/s. The correct approach involves using the equation W = F cos(Theta) (Delta y), where the force F is derived from F = ma. The participant initially calculated the force incorrectly, leading to a work value of 3900 J, which deviated significantly from the expected answer. The key takeaway is the importance of understanding the directional components of force and displacement in work calculations.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Familiarity with work-energy principles
  • Knowledge of trigonometric functions, specifically cosine
  • Basic physics concepts related to inclined planes
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  • Review the derivation of work done on an inclined plane
  • Study the application of trigonometric functions in physics problems
  • Learn about the implications of frictionless surfaces in work calculations
  • Explore advanced problems involving forces on slopes and their components
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This discussion is beneficial for physics students, educators, and anyone involved in solving mechanics problems, particularly those related to work and energy on inclined planes.

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Homework Statement


A skier of mass 67 kg is pulled up a slope by a motor-driven cable.
(a) How much work is required to pull him 40 m up a 30° slope (assumed frictionless) at a constant speed of 1.7 m/s?

m= 67kg
Delta y= 40m
Theta= 30 degrees
v= 1.7m/s


Homework Equations


I know that:

F=ma
W= F cos(Theta) (Delta y)


The Attempt at a Solution



I am solving for the force:
F=ma
F= 67 (1.7)
F= 113.9 N/m

With F solved I use my Work Equation:
W= F cos(Theta)(Delta y)
W= 113.9 (cos30) (40)
W= 3945.61 J
W= 3900 J

My answer differs from the correct answer by 10% to 100%

What am I doing wrong? Am I using the wrong equations?
 
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So you know that he has to be pulled up a distance of 40*sin(30). Work is the force times the component of the directional change projected onto the force (i.e. F \cdot \Delta s). Where does this lead you? (Hint: your directional thinking is wrong.)

Don't just use equations, know where they come from. Yes, it's hard if it's your first time, but you really can't just plug and chug and expect to get away it.
 

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