In one day a 75-kg maintain climbs ascends from the 1500-meter level on a vertic

Question

In one day a 75-kg maintain climbs ascends from the 1500-meter level on a vertical cliff to the top of 2,400 m. The next day, she dose and from the top to the base of the diff, which is at an elevation of 1350 m. Compute change it gravitational energy on the first day Compute have change of gravitational potential energy on the second day The maximum, height a human can from crouched state is what 60 cm. By how much does the gradational potential energy increase for a 72-kg person in such a jump? With what speed does the jumper leave the [Use energy principles]

Explanation / Answer

According to the given problems,

Part-A:

a) First day,

E = mgh             lets take g = 10 m/sec

E = 75kg*10*(2400-1500)

E = 675000 J

E = 675 kJ

b)Second day,

E = -mgh

E = -75*10*(2400-1350)

E = -787500 J

E = -787.5 kJ

Part-B:

a)Change in gravitational potensial energy,

Ugav = mgh

Ugav = 75*10*(0.6-0)

Ugav = 450J

b) Velocity with which he jumped, Using Conservation of energy principle,

K.E = Ugav

1/2*mv2 = 450J   as final velocity at height is zero

v = 3.464 m/s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.