Physics: A 40 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 3.7 m/s just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?

A 40 kg bear slides, from rest, 14 m down a lodgepole

Ответ:

josephbrowne9p18dit

21.01.2022

Explanation:

(a) Here, we have to take into account the earth's radius and the value of the Cavendish constant

$E_{p1}-E_{p2}=G\frac{m_{b}m_{e}}{r'}-G\frac{m_{b}m_{e}}{r}=\\6.67*10^{-11}\frac{Nm^{2}}{kg^{2}}\frac{40kg*5.9*10^{24}kg}{6371*10^{3}m}-6.67*10^{-11}\frac{Nm^{2}}{kg^{2}}\frac{40kg*5.9*10^{24}kg}{6371*10^{3}m+14m}=\\0.37J$

The change is very low due to we consider the radius of the earth as the distance to the center of mass of the earth

(b). $E_{k}=\frac{mv^{2}}{2}=\frac{(40kg)(3.7m/s)^{2}}{2}=273.8J$

(c). Is the force produced principally by the tree

$F_{f}=-\mu N$

I hope this is usefull for you

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