A 0.29-kg stone is held 1.5 m above the top edge of a water well and then droppe

Question

A 0.29-kg stone is held 1.5 m above the top edge of a water well and then dropped into it. The well has a depth of 4.8 m.

(a) Takingy = 0 at the top edge of the well, what is the gravitational potential energy of the stone–Earth system before the stone is released?
J

(b) Taking y = 0 at the top edge of the well, what is the gravitational potential energy of the stone–Earth system when it reaches the bottom of the well?
J

(c) What is the change in gravitational potential energy of the system from release to reaching the bottom of the well?
J

Explanation / Answer

here,

initial height , hi = 1.5 m

final height , hf = 4.8 m

mass , m = 0.29 kg

(a)

the gravitational potential energy of the stone–Earth system before the stone is released , PE = m * g * hi

PE = 0.29 * 9.8 * 1.5

PE = 4.26 J

(b)

the gravitational potential energy of the stone–Earth system when it reaches the bottom of the well, PEf = m * g * hf

PEf = - 0.29 * 9.8 * 4.8

PEf = - 13.64 J

(c)

the change in gravitational potential energy , delta = PEf - PEi

delta = 17.9 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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