Physics: 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?

A 40 kg girl and an 8.4 kg sled are on the frictionless

kelly1240

20.01.2022

a) $a_{sled} = 0.619\,\frac{m}{s^{2}}$ , b) $a_{girl} = 0.13\,\frac{m}{s^{2}}$ , c) $x_{girl} = 2.604\,m$

Explanation:

a) The acceleration magnitude for the sled is:

$a_{sled} = \frac{5.2\,N}{8.4\,kg }$

$a_{sled} = 0.619\,\frac{m}{s^{2}}$

b) The acceleration magnitude for the girl is:

$a_{girl} = \frac{5.2\,N}{40\,kg }$

$a_{girl} = 0.13\,\frac{m}{s^{2}}$

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

$x_{girl} = 0\,m + \frac{1}{2}\cdot (0.13\,\frac{m}{s^{2}} )\cdot t^{2}$

$x_{sled} = 15\,m-\frac{1}{2}\cdot (0.619\,\frac{m}{s^{2}} )\cdot t^{2}$

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

$0\,m + \frac{1}{2}\cdot (0.13\,\frac{m}{s^{2}} )\cdot t^{2} =15\,m-\frac{1}{2}\cdot (0.619\,\frac{m}{s^{2}} )\cdot t^{2}$

$15\,m - \frac{1}{2}\cdot (0.749\,\frac{m}{s^{2}} )\cdot t^{2} = 0$

$t \approx 6.329\,s$

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

$x_{girl} = 0\,m + \frac{1}{2}\cdot (0.13\,\frac{m}{s^{2}} )\cdot (6.329\,s)^{2}$

$x_{girl} = 2.604\,m$

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saja19

13.10.2021

All of these answers are dependent upon the specific scenario, but here are some general answers.

1. An object with a greater height will have more potential energy.

2. Potential energy can be changed into kinetic energy as an object falls. It loses height (potential energy) and gains speed (kinetic energy).

3. Depends on what scenario your class had.

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