Physics · Work, Power & Energy
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When a 7000 N elevator moves from street level to the top of a building 300 m above the street level, what is the change in gravitational potential energy?
- A
1.1 x 106 J
- B
2.1 x 106 J
- C
3.1 x 106 J
- D
4.1 x 106 J
- E
5.1 x 106 J
Change in potential energy=Work done
Change in potential energy= F.d
=7000x300
=2100000 J
2.1x106 J
As per calculations, 2.1 x 106 J is the only right option.
The change in gravitational potential energy can be
calculated using the formula:
ΔPE = m * g * Δh
where ΔPE is the change in gravitational potential energy, m
is the mass of the object, g is the acceleration due to gravity, and Δh is the
change in height.
In this case, the elevator has a weight (force) of 7000 N,
which is equivalent to its mass times the acceleration due to gravity (7000 N =
m * 9.8 m/s^2). Solving for mass, we have:
m = 7000 N / 9.8 m/s^2 ≈ 714.3 kg
The change in height is 300 m.
ΔPE = (714.3 kg) * (9.8 m/s^2) * (300 m)
ΔPE ≈ 2,090,680 Joules
Therefore, the change in gravitational potential energy of
the elevator when it moves from street level to the top of the building is
approximately 2,090,680 Joules=2.1 x 106 J
As per calculations, 2.1 x 106 J is the only right option.
As per calculations, 2.1 x 106 J is the only right option.
As per calculations, 2.1 x 106 J is the only right option.
Tagged under Physics · Work, Power & Energy · 2016