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

(a)ΔEp = 661500 J

(b)ΔEp = -771750 J

Explanation:

Conceptual analysis

We define the potential energy as that which the bodies possess by the fact of being in a certain position.That is, the body has an energy (ability to produce a job) by the fact of being at a certain height above the ground.

The formula for calculating potential energy is as follows:

Ep = m*g*h Formula (1)

Where:

Ep: It is the potential energy of the body in Joules. (J)

m: Body mass. (kg)

g: acceleration due to gravity. (m / s²)

h:height at which the body is. (m)

Change in gravitational potential energy  

We define the Change in gravitational potential energy as follows :

ΔEp= Epf- Epi Formula (2)=

Where:

ΔEp: Change in gravitational potential energy (J)

Epf : final potential energy (J)

Epi : initial potential energy (J)

(a) Change in gravitational potential energy on the first day

We apply the formula (1)

hf: final hight = 2400 m

hf: initial hight = 1500 m

We apply the formula (2):

ΔEp= Epf- Epi

We apply the formula (1):

Epf = m*g*hf

Epi = m*g*hi

ΔEp= m*g*hf - m*g*hi :We factor mg

ΔEp= m*g*(hf - *hi ) Equation (1)

ΔEp = 75*98*(2400-1500)

ΔEp =75*9.8(900)

ΔEp = 661500 J

(a) Change in gravitational potential energy on the second day

hi = 2400 m

hf = 1350 m

We apply the equation (1)

ΔEp=  m*g*(hf-hi)

ΔEp = 75*9.8*(1350-2400)

ΔEp = 75*9.8*( -1050)

ΔEp = -771750 J

a and b sometimes have a positive value

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j