Pulling a sled - find friction & force exerted on puller

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A child pulls a 9.0 kg sled through snow, accelerating it at 1.0 m/s² with a rope exerting a force of 27 N forward. According to Newton's third law, the rope must exert a force of 27 N on the child in the opposite direction of the sled's motion. The frictional force acting on the sled can be calculated using the equation F_rope - F_k = F_total, resulting in a frictional force of 18 N opposing the sled's movement. The discussion clarifies that while forces can be negative in direction, magnitudes should always be expressed as positive values. The correct abbreviation for Newton is noted as "N," not "Nt."
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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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