Math / Calculus

Question(II) A gondola can carry 20 skiers, with a total mass of up to 2250 kg . The gondola ascends at a constant speed from the base of a mountain, at 2150 m , to the summit at 3625 m . (a) How much work does the motor do in moving a full gondola up the mountain? (b) How much work does gravity do on the gondola? (c) If the motor could generate $10 \%$ more work than found in $(a)$ , what is the acceleration of the gondola?

Studdy Solution

STEP 1

1. The gondola carries a maximum total mass of $2250 \, \text{kg}$ .
2. The gondola ascends from $2150 \, \text{m}$ to $3625 \, \text{m}$ .
3. The acceleration due to gravity is $9.81 \, \text{m/s}^2$ .
4. The gondola moves at a constant speed, so initial and final kinetic energies are the same.

STEP 2

1. Calculate the work done by the motor.
2. Calculate the work done by gravity.
3. Calculate the additional work and determine the acceleration.

STEP 3

Calculate the work done by the motor:
First, find the change in height ( $\Delta h$ ):
$\Delta h = 3625 \, \text{m} - 2150 \, \text{m} = 1475 \, \text{m}$
The work done by the motor ( $W_{\text{motor}}$ ) is equal to the gravitational potential energy gained:
$W_{\text{motor}} = m \cdot g \cdot \Delta h$
Substitute the given values:
$W_{\text{motor}} = 2250 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 1475 \, \text{m}$
Calculate the result:
$W_{\text{motor}} = 32,560,875 \, \text{J}$

STEP 4

Calculate the work done by gravity:
The work done by gravity ( $W_{\text{gravity}}$ ) is the negative of the work done by the motor, as it acts in the opposite direction:
$W_{\text{gravity}} = -W_{\text{motor}}$
$W_{\text{gravity}} = -32,560,875 \, \text{J}$

STEP 5

Calculate the additional work and determine the acceleration:
First, calculate $10\%$ more work than found in (a):
$W_{\text{additional}} = 1.1 \times W_{\text{motor}}$
$W_{\text{additional}} = 1.1 \times 32,560,875 \, \text{J}$
$W_{\text{additional}} = 35,816,962.5 \, \text{J}$
The additional work provides kinetic energy for acceleration:
$W_{\text{additional}} = \frac{1}{2} m v^2$
Assuming initial velocity $v_0 = 0$ , solve for $v$ :
$v^2 = \frac{2 \times W_{\text{additional}}}{m}$
$v^2 = \frac{2 \times 35,816,962.5 \, \text{J}}{2250 \, \text{kg}}$
$v^2 = 31,847.0722 \, \text{m}^2/\text{s}^2$
$v = \sqrt{31,847.0722} \, \text{m/s}$
$v \approx 178.47 \, \text{m/s}$
Assuming constant acceleration $a$ over the distance $\Delta h$ :
$v^2 = 2a \Delta h$
$a = \frac{v^2}{2 \Delta h}$
$a = \frac{31,847.0722}{2 \times 1475}$
$a \approx 10.8 \, \text{m/s}^2$
The results are: (a) Work done by the motor: $32,560,875 \, \text{J}$ (b) Work done by gravity: $-32,560,875 \, \text{J}$ (c) Acceleration of the gondola: $10.8 \, \text{m/s}^2$

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