Question

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. then 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?

Answer 1

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

Step-by-step 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=(F)/(m)`. Now
`a_(girl)=(5.76)/(42.3)=0.14 (m/s^(2))` and
`a_(sled)=(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=(1)/(2)*a_(girl)*t^(2)` and
`15.0-x=(1)/(2)*a_(sled)*t^(2)`, solving for the time we get:
`t=\sqrt{(2x)/(a_(girl) ) }` and
`t=\sqrt{(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{(2x)/(a_(girl))} = \sqrt{(2*(15.0-x))/(a_(sled) )`. Finally we get:
`(x)/(a_(girl) )=((15.0-x))/(a_(sled) )` and replacing the values we have got:
`(x)/(0.14) =((15.0-x))/(0.73)` so
`5.33*x=15-x` so x=2.37 (meters).