Question

You are on vacation in San Francisco and decide to take a cable car to see the city. A 5800-kgkg cable car goes 260 mm up a hill inclined 17 ∘∘ above the horizontal. The system is the car and Earth. Part A Determine the change in the total energy of the system when the car moves from the bottom to the top. Ignore friction.

Answer 1

Answer:

$4.325\times10^6J$

Explanation:

Mass of the cable car, m = 5800 kg

It goes 260 m up a hill, along a slope of $\theta=17^o$

Therefore vertical elevation of the car = $260sin\theta=260sin17^o=76.0166m$

Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = $\frac{1}{2} mv^2$). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.

Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.

Hence, total change in energy = mgh = $5800\times9.81\times76.0166J=4.325\times10^6J$

where, g = acceleration due to gravity

h = height/vertical elevation

Answer 2

The change in total energy of the system is 4.32×10⁶ J

Conservation of energy:

From the law of conservation of energy, we can say that the kinetic energy of the cable car gets converted into gravitational potential energy as the cable car stops at the top and its velocity becomes zero, so the kinetic energy is zero at the top, it is used up in increasing the potential energy of the cable car as it goes up the hill.

The vertical height traveled by the cable car is:

h = 260×cos17

h = 76m

The gravitational potential energy is given by:

E = mgh

E = 5800×9.8×76 J

E = 4.32×10⁶ J

Therefore the change in total energy is 4.32×10⁶ J.

Learn more about conservation of energy:

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