purse

[logistic_guy]

A \(\displaystyle 2.0\)-\(\displaystyle \text{kg}\) purse is dropped \(\displaystyle 55 \ \text{m}\) from the top of the \(\displaystyle \text{Leaning Tower of Pisa}\) and reaches the ground with a speed of \(\displaystyle 29 \ \text{m/s}\). What was the average force of air resistance?

 

[khansaheb]

A \(\displaystyle 2.0\)-\(\displaystyle \text{kg}\) purse is dropped \(\displaystyle 55 \ \text{m}\) from the top of the \(\displaystyle \text{Leaning Tower of Pisa}\) and reaches the ground with a speed of \(\displaystyle 29 \ \text{m/s}\). What was the average force of air resistance?

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[logistic_guy]

A \(\displaystyle 2.0\)-\(\displaystyle \text{kg}\) purse is dropped \(\displaystyle 55 \ \text{m}\) from the top of the \(\displaystyle \text{Leaning Tower of Pisa}\) and reaches the ground with a speed of \(\displaystyle 29 \ \text{m/s}\). What was the average force of air resistance?

Simple Newton force equation.

\(\displaystyle mg - F_A = ma\)

We have two unknowns, but if we assume that the acceleration \(\displaystyle a\) is constant, we can calculate it from this equation:

\(\displaystyle v^2 = v^2_0 + 2ay\)

\(\displaystyle a = \frac{v^2 - v^2_0}{2y}\)

This gives:

\(\displaystyle mg - F_A = m\frac{v^2 - v^2_0}{2y}\)

Or

\(\displaystyle F_A = mg - m\frac{v^2}{2y} = m\left(g - \frac{v^2}{2y}\right)\) since \(\displaystyle v_0 = 0\).

Plug in numbers.

\(\displaystyle F_A = 2\left(9.8 - \frac{29^2}{2(55)}\right) = \textcolor{blue}{4.31 \ \text{N}}\)