Question

The summit of Mount Everest is 8848.0 m above sea
 level, making it the highest summit on Earth. In 1953, Edmund Hillary was the first person to reach the summit. Suppose upon reaching there, Hillary slid a rock from rest with a 32.0 g mass down the side of the mountain. If the rock's speed was 24.0 m/s when it was 7690.0 m above sea level, how much work was done
on the rock by air resistance?

Answer 1

$m=32.0 \;g$

$h_1=8848.0 \;m$

$h_2=7690.0\;m$

$v_1=0 \;m/s$

$v_2=24.0 \;m/s$

$g=9.81 \;m/s^2$

$W_r=m(\frac{1}2(v_2^2-v_1^2)-g(h_1-h_2))$

$W_r=(32\times 10^{-3} kg )(\frac{1}2(24^2-0^2) m^2/s^2 -9.81m/s^2(8848-7690)m)$

$W_r=(32\times 10^{-3}kg)(288-11358.4)m^2/s^2$

$W_r(32\times 10^{-3}kg)(-11060.4m^2/s^2)=-353.9J$

Answer: $W_r=-353.9 \;J$