Math  /  Calculus

Question48. (II) A 2.0kg2.0-\mathrm{kg} purse is dropped from the top of the Leaning Tower of Pisa and falls 55 m before reaching the ground with a speed of 27 m/s27 \mathrm{~m} / \mathrm{s}. What was the average force of air resistance?

Studdy Solution

STEP 1

1. The mass of the purse is 2.0kg 2.0 \, \text{kg} .
2. The purse falls a distance of 55m 55 \, \text{m} .
3. The final speed of the purse is 27m/s 27 \, \text{m/s} .
4. The acceleration due to gravity is 9.8m/s2 9.8 \, \text{m/s}^2 .
5. Air resistance is the only force doing work against gravity.

STEP 2

1. Calculate the gravitational force acting on the purse.
2. Determine the work done by gravity.
3. Calculate the kinetic energy of the purse at impact.
4. Use the work-energy principle to find the work done by air resistance.
5. Calculate the average force of air resistance.

STEP 3

Calculate the gravitational force acting on the purse:
Fgravity=mg F_{\text{gravity}} = m \cdot g =2.0kg×9.8m/s2 = 2.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 =19.6N = 19.6 \, \text{N}

STEP 4

Determine the work done by gravity:
Wgravity=Fgravity×distance W_{\text{gravity}} = F_{\text{gravity}} \times \text{distance} =19.6N×55m = 19.6 \, \text{N} \times 55 \, \text{m} =1078J = 1078 \, \text{J}

STEP 5

Calculate the kinetic energy of the purse at impact:
KE=12mv2 KE = \frac{1}{2} m v^2 =12×2.0kg×(27m/s)2 = \frac{1}{2} \times 2.0 \, \text{kg} \times (27 \, \text{m/s})^2 =729J = 729 \, \text{J}

STEP 6

Use the work-energy principle to find the work done by air resistance:
Wnet=WgravityWair resistance=KE W_{\text{net}} = W_{\text{gravity}} - W_{\text{air resistance}} = KE Wair resistance=WgravityKE W_{\text{air resistance}} = W_{\text{gravity}} - KE =1078J729J = 1078 \, \text{J} - 729 \, \text{J} =349J = 349 \, \text{J}

STEP 7

Calculate the average force of air resistance:
Wair resistance=Fair resistance×distance W_{\text{air resistance}} = F_{\text{air resistance}} \times \text{distance} Fair resistance=Wair resistancedistance F_{\text{air resistance}} = \frac{W_{\text{air resistance}}}{\text{distance}} =349J55m = \frac{349 \, \text{J}}{55 \, \text{m}} 6.35N \approx 6.35 \, \text{N}
The average force of air resistance is approximately:
6.35N \boxed{6.35 \, \text{N}}

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