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What is the change in gravitational potential energy of the apple if it falls the amount shown? E g=9.81m/s^2 and round to the nearest hundredth.

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AmbroseMaster · Tutor for 5 years

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To calculate the change in gravitational potential energy (GPE) of the apple as it falls, we can use the following formula:\[\Delta GPE = m \cdot g \cdot h\]where:- \(\Delta GPE\) is the change in gravitational potential energy,- \(m\) is the mass of the apple,- \(g\) is the acceleration due to gravity,- \(h\) is the height from which the apple falls.Given:- \(m = 0.25\) kg (mass of the apple),- \(g = 9.81 \mathrm{~m/s}^2\) (acceleration due to gravity),- \(h = 0.7\) m (height from which the apple falls).Now, let's plug these values into the formula:\[\Delta GPE = 0.25 \mathrm{~kg} \times 9.81 \mathrm{~m/s}^2 \times 0.7 \mathrm{~m}\]\[\Delta GPE = 1.7225 \mathrm{~J}\]Rounding to the nearest hundredth, the change in gravitational potential energy of the apple is:\[\Delta GPE \approx 1.72 \mathrm{~J}\]Answer: The change in gravitational potential energy of the apple is approximately 1.72 Joules.

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What is the change in gravitational potential energy of the apple if it falls the amount shown? E g=9.81m/s^2 and round to the nearest hundredth.

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