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KiRa [710]
3 years ago
8

How high off the ground is a 5.5 kg object that has a gravitational potential energy of 8500 DE=mah​

Physics
1 answer:
umka21 [38]3 years ago
7 0

Answer:

h = 157.70 meters

Explanation:

Given the following data;

Mass = 5.5 kg

Gravitational potential energy = 8500 Joules

We know that acceleration due to gravity is equal to 9.8 m/s².

To find the height of the object;

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

G.P.E = mgh

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the formula, we have;

8500 = 5.5*9.8*h

8500 = 53.9h

h = 8500/53.9

h = 157.70 m

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Answer:

The magnitude of the force is 0.7255kN

Explanation:

The elevator floor acts on the person with a force that is due to the gravitational acceleration less the downward acceleration of the elevator:

(force of floor F) = (mass of person m) x [ (grav. acceleration g) - (elevator acceleration a) ]

in other words, considering the elevator floor as a reference frame in the Earth's gravitational field, the person's weight decreases due to the downward acceleration, as follows:

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We are given the person's weight at rest, 0.9kN, from which the mass can be determined as:

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So

F = \frac{900N}{9.8 \frac{m}{s^2}}\cdot(9.8-1.9)\frac{m}{s^2}\approx 725.5N=0.7255kN

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