Question

A 40 kg girl and an 8.4 kg sled are on the frictionless ice of a frozen lake, 15 m 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 (a) the sled and (b) the girl? (c) How far from the girl's initial position do they meet?

Answer 1

a)
`a_(sled) = 0.619\,(m)/(s^(2))`, b)
`a_(girl) = 0.13\,(m)/(s^(2))`, c)
`x_(girl) = 2.604\,m`

Step-by-step explanation:

a) The acceleration magnitude for the sled is:

`a_(sled) = (5.2\,N)/(8.4\,kg )`

`a_(sled) = 0.619\,(m)/(s^(2))`

b) The acceleration magnitude for the girl is:

`a_(girl) = (5.2\,N)/(40\,kg )`

`a_(girl) = 0.13\,(m)/(s^(2))`

c) The position equations for the girl and sled are, respectively:

`x_(girl) = 0\,m + (1)/(2)\cdot (0.13\,(m)/(s^(2)) )\cdot t^(2)`

`x_(sled) = 15\,m-(1)/(2)\cdot (0.619\,(m)/(s^(2)) )\cdot t^(2)`

It is needed to know in which time the girl and sled meet each other. That is:

`0\,m + (1)/(2)\cdot (0.13\,(m)/(s^(2)) )\cdot t^(2) =15\,m-(1)/(2)\cdot (0.619\,(m)/(s^(2)) )\cdot t^(2)`

`15\,m - (1)/(2)\cdot (0.749\,(m)/(s^(2)) )\cdot t^(2) = 0`

`t \approx 6.329\,s`

The final position in comparison with the initial position of the girl is:

`x_(girl) = 0\,m + (1)/(2)\cdot (0.13\,(m)/(s^(2)) )\cdot (6.329\,s)^(2)`

`x_(girl) = 2.604\,m`