What Are the Physics Behind a Girl Pulling a Sled on Ice?

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A 40 kg girl pulls an 8.4 kg sled on frictionless ice with a force of 5.2 N. The acceleration of the sled is calculated to be 0.62 m/s², while the girl's acceleration is 0.13 m/s². To determine how far they meet from the girl's initial position, the equations of motion are set up, leading to the conclusion that they meet approximately 2.58 meters from her starting point. The calculations involve using the formula f=ma and solving for time and distance. The discussion highlights the importance of correctly applying physics equations to solve motion problems.
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I'm not sure what the rules are about posting homework problems, but I just had 2 I couldn't get around too, so here's the second one...

Homework Statement


A 40 kg girl and a 8.4 kg sled are on the frictionless ice of a frozen lake, 15m apart but connected by a rope of negligible mass. The girl exerts a horizontal 5.2 N force on the rope.
What are the acceleration magnitudes of
(a) the sledge?
(b) the girl?
(c) How far from the girl's initial position do they meet?

Homework Equations


f=ma

The Attempt at a Solution


Finding the answers to part a and b were simple..
(a) a = f/m = 5.2N/8.4kg = .62 m/s2
(b) a = f/m = 5.2N/40kg = .13 m/s2
(c) I'm having problems figuring out part C.
 
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hansel13 said:
I'm not sure what the rules are about posting homework problems, but I just had 2 I couldn't get around too, so here's the second one...

Homework Statement


A 40 kg girl and a 8.4 kg sled are on the frictionless ice of a frozen lake, 15m apart but connected by a rope of negligible mass. The girl exerts a horizontal 5.2 N force on the rope.
What are the acceleration magnitudes of
(a) the sledge?
(b) the girl?
(c) How far from the girl's initial position do they meet?

Homework Equations


f=ma

The Attempt at a Solution


Finding the answers to part a and b were simple..
(a) a = f/m = 5.2N/8.4kg = .62 m/s2
(b) a = f/m = 5.2N/40kg = .13 m/s2
(c) I'm having problems figuring out part C.

Write equations for their positions
XG + XS = 15
XG = 1/2*ag*t2
XS = 1/2*as*t2

Solve noting that t2 is the same when they meet
 


1/2*ag*t2+1/2*as*t2= 15

t2 = 15 / (1/2*ag +1/2*as)
t2 = 15 / (1/2*.62 +1/2*.13)
t2 = 40
t = 6.3m

That doesn't seem right, I'm lost.
 


hansel13 said:
1/2*ag*t2+1/2*as*t2= 15

t2 = 15 / (1/2*ag +1/2*as)
t2 = 15 / (1/2*.62 +1/2*.13)
t2 = 40
t = 6.3m

That doesn't seem right, I'm lost.

Not quite. That gives you seconds.

XG = 1/2*ag*t2 = 1/2*(.13)*(6.3)2 = 2.58m
 


LowlyPion said:
Not quite. That gives you seconds.

XG = 1/2*ag*t2 = 1/2*(.13)*(6.3)2 = 2.58m

ahh. Gotcha, thanks.
 
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