Pulling a sled - find friction & force exerted on puller

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SUMMARY

The discussion focuses on the physics of pulling a sled with a mass of 9.0 kg, where the sled accelerates at 1.0 m/s² under the influence of a rope exerting a force of 27 N. According to Newton's Third Law, the force exerted by the rope on the child is also 27 N but in the opposite direction. The frictional force opposing the sled's motion is calculated to be 18 N, derived from the equation Frope - Fk = ma, confirming that the sled experiences friction while moving through the snow.

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  • Understanding of Newton's Laws of Motion
  • Basic knowledge of force and acceleration equations (F=ma)
  • Concept of frictional forces in physics
  • Familiarity with units of measurement, specifically Newtons (N)
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  • Investigate the role of tension in ropes and its applications in physics problems
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of forces, particularly in scenarios involving friction and tension in pulling systems.

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1. A child pulls a sled with mass 9.0kg through the snow(note: no coefficient of friction given, but there _is_ friction between sled & snow). The sled accelerates at 1.0 m/s2. The rope exerts 27Nt on the sled in the forward direction.

a) what force must the rope exert on the child (magnitude & direction)?

b) What is the magnitude of the frictional force on the sled (hint: it's in the opposite direction of the sled's motion)?





Homework Equations



F=ma

The Attempt at a Solution



a) 27 Nt, due to Newton's 3rd law, in the opposite direction of the child/sled's motion.


b)
Frope - Fk = Ftotal = ma

Frope = Fk + ma

27 = Fk + (9 * 1)

18Nt = Fk


are my answers correct? Should their signs be negative, because they are in the opposite direction of the sled's motion? Have I made any glaring omissions regarding the rope's tension?
 
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Looks good to me. Magnitudes are always positive. (The abbreviation for Newton is N, not Nt.)
 

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