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Alenkasestr [34]
3 years ago
8

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

n the girl pulls on the rope, exerting a 5.76 N force on the sled, pulling it toward her. how far from the girl's original position do they meet?
Physics
1 answer:
ycow [4]3 years ago
5 0

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: F=m*a with this we can get both accelerations; solving for acceleration a=\frac{F}{m}. Now a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2}) anda_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2}). Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, x=\frac{1}{2}*a_{girl}*t^{2} and15.0-x=\frac{1}{2}*a_{sled}*t^{2}, solving for the time we get:t=\sqrt{\frac{2x}{a_{girl} } } and t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } } with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get:\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }. Finally we get:\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} } and replacing the values we have got:\frac{x}{0.14} =\frac{(15.0-x)}{0.73} so 5.33*x=15-x so x=2.37 (meters).

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