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An elevator car has a mass of 750 kg, and its three passengers have a combined mass of 135 kg. If the elevator and its passengers ride to the ground floor, 20.7 m below, find the change in gravitational potential energy of the car and its passengers. Show all your work.

Sagot :

whitneytr12

Answer:

The change in gravitational potential energy is -1.80x10⁵ J.

Explanation:

The change in gravitational potential energy is given by:

[tex] \Delta E_{p} = E_{p_{f}} - E_{p_{i}} [/tex]

[tex] \Delta E_{p} = mgh_{f} - mgh_{i} [/tex]

Where:

"i" is for final and "f" for final

m: is the mass

g: is the gravity = 9.81 m/s²

h: is the height

For the car and the passengers we have:

[tex] \Delta E_{p} = m_{T}g(h_{f} - h_{i}) = (750 kg + 135 kg)9.81 m/s^{2}(0 - 20.7 m) = -1.80 \cdot 10^{5} J [/tex]      

The minus sign is because when the elevator car and the passengers are up they have a bigger gravitational potential energy than when they are in the ground.

Therefore, the change in gravitational potential energy is -1.80x10⁵ J.

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