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A 2-kg block is thrown upward from a point 20 m above the Earth's surface. At what height above Earth's surface will the gravitational potential energy of the Earth-block system have increased by 500 J?

User Anya Samadi
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2 Answers

14 votes
14 votes

Final answer:

The new height at which the gravitational potential energy has increased by 500 J can be found using the formula for gravitational potential energy and by rearranging it to solve for the new height.

Step-by-step explanation:

To determine at what height above Earth's surface the gravitational potential energy of the Earth-block system will have increased by 500 J for a 2-kg block thrown upward, we use the formula for gravitational potential energy (GPE), which is GPE = mgh, where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s² on Earth), and h is the height above the reference point. First, we calculate the initial GPE at 20 m height by plugging in the values we have (GPE = 2 kg × 9.8 m/s² × 20 m), then we add the 500 J to find the new total GPE. Afterward, we solve for the new height h by rearranging the GPE formula and dividing the total GPE by the product of the mass and gravitational acceleration.

User Kokovin Vladislav
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3.4k points
13 votes
13 votes

Answer:

45.5 m

Step-by-step explanation:

m = 2 kg, h = 20 m, E = 500 J, radius of earth = R, mass of earth = M

find the new height H

at h, the potential energy = -GMm/(R + h)

at H, the potential energy = -GMm/(R + H)

increase of the potential energy

= [-GMm/(R + H)] - [-GMm(R + h)]

= GMm[1/(R + h) - 1/(R + H)] = E

1/(R + h) - 1/(R + H) = E/(GMm)

(H - h)/[(R + H)(R + h)] = E/(GMm)

R + h ≈ R, R + H ≈ R

so (H - h)/R² = E/(GMm)

H - h = ER²/(GMm)

note GM/R² = g = 9.81 m/s²

so H - h = E/(mg)

H = h + E/(mg) = 20 + 500/(2*9.81) = 45.5 m

User Rossella
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