In one day, a 75 kgmountain climber ascends from the 1500mlevel on a vertical cliff to the top at 2400m. The next day, she descends from the top to the base of the cliff, which is at an elevation of 1350m. What is her change in gravitational potential energy (a) on the first day and (b) on the second day?

The given data can be listed below as:

• The mass of the mountain climber is, m = 75 kg.
• In the first case, the height reached by the mountain climber initially is, $h_1=1500m$.
• In the first case, the height reached by the mountain climber finally is, $h_2=2400m$.
• In the second case, the height reached by the mountain climber initially is, $h_3=2400m$.
• In the second case, the height reached by the mountain climber finally is, $h_4=1350m$.

The gravitational potential energy is described as the energy that an object possesses with the change in the position of that object. The gravitational potential energy is the product of the mass of an object, acceleration due to gravity and height acquired by that object.

The equation of the gravitational potential energy is expressed as:

$\mathrm{U}=\mathrm{mg}({\mathrm{h}}_2-{\mathrm{h}}_1)$

Here, U is the gravitational potential energy, m is the mass of the mountain climber, g is the acceleration due to gravity,$h_2$ is the height reached by the mountain climber finally in the first case and$h_1$ is the height reached by the mountain climber initially in the first case.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{U}=\left(75\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2400\mathrm{m}-1500\mathrm{m}\right) \\ =\left(735\mathrm{kg}\times \mathrm{m}/{\mathrm{s}}^2\right)\left(900\mathrm{m}\right) \\ =6.6\times {10}^5\mathrm{kg}\times {\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}\times {\mathrm{m}}^2/{\mathrm{s}}^2} \\ =6.6\times {10}^5\mathrm{J} \end{gathered}$$

Thus, the change in gravitational potential energy on the first day is $6.6\times {10}^5\mathrm{J}$.

The equation of the gravitational potential energy is expressed as:

$\mathrm{U}=\mathrm{mg}({\mathrm{h}}_4-{\mathrm{h}}_3)$

Here, U is the gravitational potential energy, m is the mass of the mountain climber, g is the acceleration due to gravity,$h_4$ is the height reached by the mountain climber finally in the second case and$h_3$ is the height reached by the mountain climber initially in the second case.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{U}=\left(75\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1350\mathrm{m}-2400\mathrm{m}\right) \\ =\left(735\mathrm{kg}\times \mathrm{m}/{\mathrm{s}}^2\right)\left(-1050\mathrm{m}\right) \\ =-7.7\times {10}^5\mathrm{kg}\times {\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}\times {\mathrm{m}}^2/{\mathrm{s}}^2} \\ =-7.7\times {10}^5\mathrm{J} \end{gathered}$$

Thus, the change in gravitational potential energy on the second day is $-7.7\times {10}^5\mathrm{J}$.