Newton's Law: sliding down a slope

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SUMMARY

The discussion focuses on calculating the work done by friction when a snow sled, with a total mass of 80.0 kg, slides down a slope at a constant speed. The slope angle is 60.0°, and the coefficient of kinetic friction is 0.290. The user derived the forces involved, finding the weight force (Fw) to be 541 N and the friction force (Ff) to be 124 N. The net work done by friction can be calculated using the formula W = Ff * d, where d is the distance slid down the slope, which is 14.0 m.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Basic knowledge of forces, including friction and tension
  • Familiarity with trigonometric functions in physics
  • Ability to apply work-energy principles
NEXT STEPS
  • Calculate the work done by friction using the formula W = Ff * d
  • Explore the implications of constant speed on net forces in inclined planes
  • Investigate the role of the coefficient of kinetic friction in motion analysis
  • Learn about tension forces in systems involving inclined planes
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Physics students, educators, and anyone interested in understanding the dynamics of motion on inclined planes, particularly in the context of friction and work calculations.

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A snow sled, with total mass 80.0 kg, is lowered at constant speed down a slope of angle 60.0° with respect to the horizontal, for a distance d = 14.0 m. The coefficient of kinetic friction between the sled and the snow is 0.290.

Note: g = 9.80 m s–2.

1. What is the work done by the friction force when the sled slides a distance d down the inclined slope?
Equations tried using:
Fnet = F1 + F2 + F3
Ffr = µf(mgcos theta)
Fnet = Wsin theta - F friction
i found the Fw to be 541N and the Ff to be 124N but then i seem to get lost on how to get tension force to get the net work done by friction, i used the formula Fnet = Fw-T-Ff = 0
Fw - Ff = T = 416N then got stuck
 
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I do not see an equation for work - which is what you are supposed to find. What is it?
 

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